Common MySQL Queries
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Entries in italics new or updated in the last monthBasic aggregation
This is the simplest grouping query pattern. For column foo, display the first (smallest), last (largest) or average value of column bar:
SELECT foo, MIN(bar) AS bar
FROM tbl
GROUP BY foo
Return the highest bar value for each foo, ordering top to bottom by that value:
SELECT foo, MAX(bar) AS Count
FROM tbl
GROUP BY foo
ORDER BY Count DESC;
Ditto for AVG(), COUNT() etc. The pattern is easily extended for multiple grouping column expressions.
MySQL introduced the SQL extension GROUP_CONCAT(), which makes short work of listing items in groups. For example, given a table of suppliers and the parts they make ...
CREATE TABLE supparts(supID char(2),partID char(2));
INSERT INTO supparts VALUES
('s1','p1'),('s1','p2'),('s1','p3'),('s1','p4'),('s1','p5'),('s1','p6'),
('s2','p1'),('s2','p2'),('s3','p2'),('s4','p2'),('s4','p4'),('s4','p5');
list suppliers for each part:
SELECT partID,GROUP_CONCAT(supID ORDER BY supID) AS Suppliers
FROM supparts
GROUP BY partID;
+--------+-------------+
| partID | Suppliers |
+--------+-------------+
| p1 | s1,s2 |
| p2 | s1,s2,s3,s4 |
| p3 | s1 |
| p4 | s1,s4 |
| p5 | s1,s4 |
| p6 | s1 |
+--------+-------------+
Aggregate across columns
You track squash court bookings and fees. A court booking has one fee, but it has two member references, and those can be the same if one member has brought a guest. For each booking row, the fee is to be split between the two members. What query correctly aggregates fees including cases where the two members of record are the same?
DROP TABLE IF EXISTS bookings;
CREATE TABLE bookings (
court_id int(11) NOT NULL,
member1 int(11) NOT NULL,
member2 int(11) NOT NULL,
time timestamp NOT NULL,
fee decimal(5,2) NOT NULL
);
INSERT INTO bookings ( court_id , member1 , member2 , time , fee )
VALUES
(1, 1000, 1001, '2009-09-09 15:49:38', 3.00),
(2, 1000, 1000, '2009-09-08 15:50:04', 3.00);
For this data the correct result is
member fees
1000 4.50
1001 1.50
An efficient solution, posted by 'laptop alias' on a MySQL forum:
SELECT member, ROUND(SUM(fee/2),2) AS total
FROM (
SELECT member1 AS member, fee FROM bookings
UNION ALL
SELECT member2, fee FROM bookings
) AS tmp
GROUP BY member;
Last updated 09 Sep 2009
Aggregates across multiple joins
Given a parent table and two child tables, a query which sums values in both child tables, grouping on a parent table column, returns sums that are exactly twice as large as they should be. In this example from the MySQL General Discussion list:
DROP TABLE IF EXISTS packageItem,packageCredit,packageItemTax;
CREATE TABLE packageItem (
packageItemID INT,
packageItemName CHAR(20),
packageItemPrice DECIMAL(10,2)
);
INSERT INTO packageItem VALUES(1,'Delta Hotel',100.00);
CREATE TABLE packageCredit (
packageCreditID INT,
packageCreditItemID INT,
packageItemType CHAR(10),
packageCreditAmount DECIMAL(10,2)
);
INSERT INTO packageCredit VALUES
(1,1,'Deposit',25.00),
(2,1,'Balance',92.00);
CREATE TABLE packageItemTax (
packageItemTaxID INT,
packageItemTaxItemID INT,
packageItemTaxName CHAR(5),
packageItemTaxAmount DECIMAL(10,2)
);
INSERT INTO packageItemTax VALUES
(1,1,'GST',7.00),
(2,1,'HST',10.00);
The query ...
SELECT
packageItemID AS Item,
SUM(packageItemPrice) AS Price,
SUM(packageItemTaxAmount) AS Tax,
SUM(packageCreditAmount) AS Credit
FROM packageItem
LEFT JOIN packageCredit ON packageItemID=packageCreditItemID
LEFT JOIN packageItemTax ON packageItemTaxItemID=packageItemID
GROUP BY packageItemID
ORDER BY packageItemID;
returns this incorrect result ...
+------+--------+-------+--------+
| Item | Price | Tax | Credit |
+------+--------+-------+--------+
| 1 | 400.00 | 34.00 | 234.00 |
+------+--------+-------+--------+
With three child tables, the sums are tripled. Why? The query aggregates across each join. How then to get the correct results? One way is with correlated subqueries:
SELECT
packageItemID AS Item,
SUM(packageItemPrice) AS Price,
( SELECT SUM(c.packageCreditAmount)
FROM packageCredit c
WHERE c.packageCreditItemID = packageItemID
) AS Credit,
( SELECT SUM(t.packageItemTaxAmount)
FROM packageItemTax t
WHERE t.packageItemTaxItemID = packageItemID
) AS Tax
FROM packageItem
GROUP BY packageItemID;
+------+--------+--------+-------+
| Item | Price | Credit | Tax |
+------+--------+--------+-------+
| 1 | 100.00 | 117.00 | 17.00 |
+------+--------+--------+-------+
Moving the subquery logic to the JOIN level may speed up performance considerably:
SELECT
i.packageItemID AS Item,
SUM(i.packageItemPrice) AS Price,
c.Credit,
t.Tax
FROM packageItem i
JOIN (
SELECT packageCreditItemID, SUM(packageCreditAmount) AS Credit
FROM packageCredit
) c ON i.packageItemID = c.packageCreditItemID
JOIN (
SELECT packageItemTaxItemID, SUM(t.packageItemTaxAmount) AS Tax
FROM packageItemTax t
) t ON i.packageItemID = t.packageItemTaxItemID
GROUP BY packageItemID;
If subqueries are unavailable or too slow, replace them with temp tables.
Aggregates excluding leaders
You have a table of grouped ranks ...
DROP TABLE IF EXISTS grps,ranks;
CREATE TABLE grps (grp int);
INSERT INTO grps VALUES(1),(2),(3),(4);
CREATE TABLE ranks(grp int,rank int);
INSERT INTO ranks VALUES(1, 4 ),(1, 7 ),(1, 9 ),(2, 2 ),(2, 3 ),(2, 5 ),(2, 6 ),(2, 8 ),(3, 1 ),(4,11 ),(4,12 ),(4,13 );
and you wish to list ranks by group omitting the leading rank in each group. The simplest query for group leaders is ...
SELECT grp, MIN(rank) as top
FROM ranks r2
GROUP BY grp
+------+------+
| grp | top |
+------+------+
| 1 | 4 |
| 2 | 2 |
| 3 | 1 |
| 4 | 11 |
+------+------+
The simplest way to get a result that omits these is an exclusion join from the ranks table to the above result:
SELECT r1.grp, r1.rank
FROM ranks r1
LEFT JOIN (
SELECT grp, MIN(rank) as top
FROM ranks r2
GROUP BY grp
) AS r2 ON r1.grp=r2.grp AND r1.rank = r2.top
WHERE r2.grp IS NULL
ORDER BY grp, rank;
+------+------+
| grp | rank |
+------+------+
| 1 | 7 |
| 1 | 9 |
| 2 | 3 |
| 2 | 5 |
| 2 | 6 |
| 2 | 8 |
| 4 | 12 |
| 4 | 13 |
+------+------+
Aggregates of specified size
Find the values of a table column c1 for which there are a specified number of listed values in another column c2.
To get an overview of the values of c2 for each value of c1:
SELECT
c1,
GROUP_CONCAT(c2 ORDER BY c2) AS 'C2 values'
FROM table
GROUP BY c1;
To retrieve a list of c1 values for which there exist specific values in another column c2, you need an IN clause specifying the c2 values and a HAVING clause specifying the required number of different items in the list ...
SELECT c1
FROM table
WHERE c2 IN (1,2,3,4)
GROUP BY c1
HAVING COUNT(DISTINCT c2)=4;
This is easy to generalise to multiple column expressions, and a HAVING clause specifying any number of items from the IN list.
All X for which all Y are Z
You have an election database with tables for candidates, parties and districts. A candidate belongs to one party; a district may have any number of candidates:
DROP TABLE IF EXISTS parties,districts,candidates;
CREATE TABLE parties (
party char(12) NOT NULL,
PRIMARY KEY (party)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO parties VALUES ('Conservative'),('Liberal'),('Socialist'),('Green'),('Libertarian');
CREATE TABLE districts (
district char(10) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO districts VALUES ('Essex'),('Malton'),('Riverdale'),('Guelph'),('Halton');
CREATE TABLE candidates (
id int(11) NOT NULL,
name char(10) DEFAULT NULL,
district char(10) DEFAULT NULL,
party char(10) DEFAULT NULL,
PRIMARY KEY (id)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO candidates VALUES
(1,'Anne Jones','Essex','Liberal'),(2,'Mary Smith','Malton','Liberal'),
(3,'Sara Black','Riverdale','Liberal'),(4,'Paul Jones','Essex','Socialist'),
(5,'Ed White','Essex','Conservative'),(6,'Jim Kelly','Malton','Liberal'),
(7,'Fred Price','Riverdale','Socialist'),(8,'Bill Green','Guelph','Green'),
(9,'Garth Adams','Halton','Libertarian'),(10,'Sam Adams','Guelph','Liberal'),
(11,'Jill Mackay','Halton','Liberal');
What query shows which parties have candidates in all districts? The simplest solution is to aggregate on party from a join of candidates to districts, and condition the result on each party having a rowcount at least equal to the district count:
SELECT party
FROM candidates
INNER JOIN districts USING (district)
GROUP BY party
HAVING COUNT(party) >= (SELECT COUNT(*) FROM districts);
+---------+
| party |
+---------+
| Liberal |
+---------+
The
query pattern is: given a table of x values in table X, a table of y values in table Y, and a table of x-y pairs in table Z, use the following query to find all X.x which are paired with every Y.y in Z:
SELECT x
FROM Y JOIN Z USING (y)
GROUP BY x
HAVING COUNT(x) >= (SELECT COUNT(*) FROM y);
For a very different solution see
relational division.
Avoiding repeat aggregation
In a good introductory
tutorial on MySQL subqueries, Jeremy Cole developed a triply nested query to retrieve the name, population, head of state, and number of official languages in countries with the most official languages. The query uses two tables from the MySQL
world database:
CREATE TABLE country (
Code char(3) NOT NULL DEFAULT '' PRIMARY KEY,
Name char(52) NOT NULL DEFAULT '',
Population int(11) NOT NULL DEFAULT '0',
HeadOfState char(60) DEFAULT NULL,
... other columns ...
);
CREATE TABLE countrylanguage (
CountryCode char(3) NOT NULL DEFAULT '' PRIMARY KEY,
Language char(30) NOT NULL DEFAULT '',
IsOfficial enum('T','F') NOT NULL DEFAULT 'F',
Percentage float(4,1) NOT NULL DEFAULT '0.0'
);
The query needs to aggregate language counts by country twice: once for all language counts by country, and once again to identify countries with the highest number of languages:
SELECT name, population, headofstate, top.num
FROM Country
JOIN (
SELECT countrycode, COUNT(*) AS num
FROM CountryLanguage
WHERE isofficial='T'
GROUP BY countrycode
HAVING num = (
SELECT MAX(summary.nr_official_languages)
FROM (
SELECT countrycode, COUNT(*) AS nr_official_languages
FROM CountryLanguage
WHERE isofficial='T'
GROUP BY countrycode
) AS summary
)
) as top ON Country.code=top.countrycode;
+--------------+------------+-------------+-----+
| name | population | headofstate | num |
+--------------+------------+-------------+-----+
| Switzerland | 7160400 | Adolf Ogi | 4 |
| South Africa | 40377000 | Thabo Mbeki | 4 |
+--------------+------------+-------------+-----+
In addition, one of the nested subqueries is buried in a
HAVING clause. This is fine with small tables, but if the table being aggregated is very large and the aggregation is complex, performance may be unsatisfactory. Substituting a temporary table for the double nesting can improve performance in two ways:
- the aggregation needs to be done just once
- we can use an exclusion join, which is usually faster than a
HAVING clause, to find countries with the maximum counts:
DROP TABLE IF EXISTS top;
CREATE TABLE top ENGINE=MEMORY
SELECT countrycode, COUNT(*) AS num
FROM CountryLanguage l1
WHERE isofficial='T'
GROUP BY countrycode;
SELECT name,population,headofstate,t3.num
FROM country c
JOIN (
SELECT t1.countrycode, t1.num
FROM top t1
LEFT JOIN top t2 ON t1.num < t2.num
WHERE t2.countrycode IS NULL
) AS t3 ON c.code=t3.countrycode;
+--------------+------------+-------------+-----+
| name | population | headofstate | num |
+--------------+------------+-------------+-----+
| Switzerland | 7160400 | Adolf Ogi | 4 |
| South Africa | 40377000 | Thabo Mbeki | 4 |
+--------------+------------+-------------+-----+
DROP TABLE top;
You notice that we haven't actually used a
TEMPORARY table? Indeed we haven't, because of the MySQL limitation that temporary tables cannot be referenced multiple times in a query. Until that's lifted, we get almost as much speed improvement from using a
MEMORY table as a temporary table.
Cascading aggregates
When you have parent-child-grandchild tables, eg
companies, users, actions, and your query requirement is for per-parent aggregates from the child table and per-child aggregates from the grandchild table, then cascading joins yield spuriously multiplied counts, and correlated subqueries fail because the second correlated subquery cannot find a visible joining column.
One solution is to use derived tables. Assuming ...
DROP TABLE IF EXISTS companies,users,actions;
CREATE TABLE companies (id int, name char(10));
INSERT INTO COMPANIES VALUES(1,'abc ltd'),(2,'xyz inc');
CREATE TABLE users (id INT,companyid INT);
INSERT INTO users VALUES(1,1),(2,1),(3,1),(4,2),(5,2);
CREATE TABLE actions (id INT, userid INT, date DATE);
INSERT INTO actions VALUES
( 1, 1, '2009-1-2'),( 2, 1, '2009-1-3'),( 3, 2, '2009-1-4'),( 4, 2, '2009-1-5'),( 5, 3, '2009-1-6'),
( 6, 3, '2009-1-7'),( 7, 4, '2009-1-8'),( 8, 5, '2009-1-9'),( 9, 5, '2009-1-9'),(10, 5, '2009-1-9');
then...
- Join
companies and users once to establish a derived company-user table.
- Join them a second time, this time aggregating on
users.id to retrieve user counts per company.
- Join the first derived table to the actions table, aggregating on
actions.id to retrieve actions per user per company:
Here is the SQL:
SELECT cu1.cname as CpyName, cu2.uCnt as Users, ua.uid as UserNo, ua.aCnt as Actions, ua.Latest
FROM (
SELECT c.id AS cid, c.name AS cname, u1.id AS uid
FROM companies c
INNER JOIN users u1 ON u1.companyid=c.id
) AS cu1
INNER JOIN (
SELECT c.id AS cid, COUNT(u2.id) AS uCnt
FROM companies c
INNER JOIN users u2 ON u2.companyid=c.id
GROUP BY c.id
) AS cu2 ON cu1.cid=cu2.cid
INNER JOIN (
SELECT u3.id AS uid, COUNT(a.id) AS aCnt, MAX(a.date) AS latest
FROM users u3
INNER JOIN actions a ON u3.id=a.userid
GROUP BY u3.id
) AS ua ON ua.uid=cu1.uid;
+---------+-------+--------+---------+------------+
| CpyName | Users | UserNo | Actions | Latest |
+---------+-------+--------+---------+------------+
| abc ltd | 3 | 1 | 2 | 2009-01-03 |
| abc ltd | 3 | 2 | 2 | 2009-01-05 |
| abc ltd | 3 | 3 | 2 | 2009-01-07 |
| xyz inc | 2 | 4 | 1 | 2009-01-08 |
| xyz inc | 2 | 5 | 3 | 2009-01-09 |
+---------+-------+--------+---------+------------+
Cross-aggregates
Given the table authorbook(authid INT, bookid INT), what query finds the books who have authors with more than one book in the table?
Even one level of recursion can induce a mild trance. Escape the trance by taking the problem one step at a time. First write the query that finds the authors with multiple books. Then join an outer query to that on authorid, and have the outer query select bookid:
SELECT a1.bookid
FROM authorbook a1
INNER JOIN (
SELECT authid,count(bookid)
FROM authorbook a2
GROUP BY authid
HAVING COUNT(bookid)>1
) AS a3 ON a1.authid=a3.authid;
Group data by datetime periods
To group rows by a time period whose length in minutes divides evenly into 60, use this formula:
GROUP BY ((60/periodMinutes) * HOUR( thistime ) + FLOOR( MINUTE( thistime ) / periodMinutes ))
where thistime is the TIME column and periodMinutes is the period length in minutes. So to group by 15-min periods, write ...
SELECT ...
GROUP BY ( 4 * HOUR( thistime ) + FLOOR( MINUTE( thistime ) / 15 ))
...
A simpler application of the same logic works for hours. For example, list data by three-hour periods:
DROP TABLE IF EXISTS t;
CREATE TABLE t(t time,i int);
INSERT INTO t VALUES('01:01:01',1),('02:02:02',2),('05:05:05',5);
SELECT FLOOR(HOUR(t)/3) AS period, GROUP_CONCAT(i) AS i
FROM t
GROUP BY period;
+--------+------+
| period | i |
+--------+------+
| 0 | 1,2 |
| 1 | 5 |
+--------+------+
And the same logic works for months ...
GROUP BY ((12/periodMonths) * YEAR( thisdate ) + FLOOR( MONTH( thisdate ) / periodMonths ))
It could be made to work for weeks with a function that maps the results of WEEK() to the range 1...52.
When the desired grouping period is a value returned by a MySQL date-time function, matters become simpler: just group by the desired value. Thus to group by weeks, write ..
SELECT ...
GROUP BY WEEK( datecol)
...
You can also Group By an expression like
CEIL( TIME_TO_SEC( TIMEDIFF( timestamp1, timestamp2 )) / (60*60) )
modifying the denominator to suit.
If there is no expression invoking a MySQL date-time function that returns the desired grouping period, you will need to write your own stored function.
League table
Here is a simple soccer league table setup that was developed in the MySQL Forum by J Williams and a contributor named "Laptop Alias". The teams table tracks team ID and name, the games table tracks home and away team IDs and goal totals for each game. The query for standings is built by aggregating a UNION of home team and away team game results:
DROP TABLE IF EXISTS teams, games;
CREATE TABLE teams(id int primary key auto_increment,tname char(32));
CREATE TABLE games(id int primary key auto_increment, date datetime,
hteam int, ateam int, hscore tinyint,ascore tinyint);
INSERT INTO teams VALUES(1,'Wanderers'),(2,'Spurs'),(3,'Celtics'),(4,'Saxons');
INSERT INTO games VALUES
(1,'2008-1-1 20:00:00',1,2,1,0),(2,'2008-1-1 20:00:00',3,4,0,2),
(3,'2008-1-8 20:00:00',1,3,1,1),(4,'2008-1-8 20:00:00',2,4,2,1);
SELECT * FROM teams;
+----+-----------+
| id | tname |
+----+-----------+
| 1 | Wanderers |
| 2 | Spurs |
| 3 | Celtics |
| 4 | Saxons |
+----+-----------+
SELECT * FROM games;
+----+---------------------+-------+-------+--------+--------+
| id | date | hteam | ateam | hscore | ascore |
+----+---------------------+-------+-------+--------+--------+
| 1 | 2008-01-01 20:00:00 | 1 | 2 | 1 | 0 |
| 2 | 2008-01-01 20:00:00 | 3 | 4 | 0 | 2 |
| 3 | 2008-01-08 20:00:00 | 1 | 3 | 1 | 1 |
| 4 | 2008-01-08 20:00:00 | 2 | 4 | 2 | 1 |
+----+---------------------+-------+-------+--------+--------+
-- Standings query:
SELECT
tname AS Team, Sum(P) AS P,Sum(W) AS W,Sum(D) AS D,Sum(L) AS L,
SUM(F) as F,SUM(A) AS A,SUM(GD) AS GD,SUM(Pts) AS Pts
FROM(
SELECT
hteam Team,
1 P,
IF(hscore > ascore,1,0) W,
IF(hscore = ascore,1,0) D,
IF(hscore < ascore,1,0) L,
hscore F,
ascore A,
hscore-ascore GD,
CASE WHEN hscore > ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END PTS
FROM games
UNION ALL
SELECT
ateam,
1,
IF(hscore < ascore,1,0),
IF(hscore = ascore,1,0),
IF(hscore > ascore,1,0),
ascore,
hscore,
ascore-hscore GD,
CASE WHEN hscore < ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END
FROM games
) as tot
JOIN teams t ON tot.Team=t.id
GROUP BY Team
ORDER BY SUM(Pts) DESC ;
+-----------+------+------+------+------+------+------+------+------+
| Team | P | W | D | L | F | A | GD | Pts |
+-----------+------+------+------+------+------+------+------+------+
| Wanderers | 2 | 1 | 1 | 0 | 2 | 1 | 1 | 4 |
| Spurs | 2 | 1 | 0 | 1 | 2 | 2 | 0 | 3 |
| Saxons | 2 | 1 | 0 | 1 | 3 | 2 | 1 | 3 |
| Celtics | 2 | 0 | 1 | 1 | 1 | 3 | -2 | 1 |
+-----------+------+------+------+------+------+------+------+------+
Monthly expenses
You have four tables to track revenue and expenses—bankaccount, cash, accountitem, accountcategory:
drop table if exists accountitem,accountcategory,bankaccount,cash;
create table accountitem(
itemid int primary key auto_increment,itemname char(32),itemcatid int
);
create table accountcategory(
categoryid int primary key auto_increment,categoryname char(32),isexpense bool
);
create table bankaccount(
id int auto_increment primary key,amount decimal(12,2),itemid int,entrydate date
);
create table cash(
id int auto_increment primary key,amount decimal(12,2),itemid int,date date
);
You need monthly expense totals. The solution uses two query patterns—
- sum across a join of cash and bank tables for totals
- pivot on month to break out monthly totals
select
if(month=1, m.Amt+n.Amt,0 ) As `Jan`, -- pivot by month
if(month=2, m.Amt+n.Amt,0 ) As `Feb`,
if(month=3, m.Amt+n.Amt,0 ) As `Mar`,
if(month=4, m.Amt+n.Amt,0 ) As `Apr`,
if(month=5, m.Amt+n.Amt,0 ) As `May`,
if(month=6, m.Amt+n.Amt,0 ) As `Jun`,
if(month=7, m.Amt+n.Amt,0 ) As `Jul`,
if(month=8, m.Amt+n.Amt,0 ) As `Aug`,
if(month=9, m.Amt+n.Amt,0 ) As `Sep`,
if(month=10,m.Amt+n.Amt,0 ) As `Oct`,
if(month=11,m.Amt+n.Amt,0 ) As `Nov`,
if(month=12,m.Amt+n.Amt,0 ) As `Dec`
from (
select Month(a.Date) As Month, sum(a.Amount) as Amt
from cash a
join accountitem b on a.itemid=b.itemid
join accountcategory c on b.ItemCatID=c.CategoryID
where c.IsExpense=1
Group By Month
) As m join ( -- join cash & bank data
select Month(a.EntryDate) As Month, sum(a.Amount ) AS Amt
from BankAccount a
join accountitem b on a.itemid=b.itemid
join accountcategory c on b.ItemCatID=c.CategoryID
where c.IsExpense=1
Group By Month
) As n using(Month);
Nested aggregation
Employee sales commission rates increase with sales totals according to specified bands of sales total amounts—like a graduated income tax in reverse. To compute total commissions due each employee, we need to aggregate twice: first to get sales per employee, then to get commissions per employee:
DROP TABLE IF EXISTS sales, commissions;
CREATE TABLE sales(employeeID int,sales int);
INSERT INTO sales VALUES(1,2),(1,5),(1,7),(2,9),(2,15),(2,12);
SELECT * FROM sales;
+------------+-------+
| employeeID | sales |
+------------+-------+
| 1 | 2 |
| 1 | 5 |
| 1 | 7 |
| 2 | 9 |
| 2 | 15 |
| 2 | 12 |
+------------+-------+
CREATE TABLE commissions(
comstart DECIMAL(6,2),
commend DECIMAL(6,2),
comfactor DECIMAL(6,2),
pct INT
);
INSERT INTO commissions VALUES
(1.00,10.00,0.10,10),(11.00,20.00,0.20,20),(21.00,30.00,0.30,30),(31.00,40.00,0.40,40);
SELECT * FROM commissions;
+----------+---------+-----------+------+
| comstart | commend | comfactor | pct |
+----------+---------+-----------+------+
| 1.00 | 10.00 | 0.10 | 10 |
| 11.00 | 20.00 | 0.20 | 20 |
| 21.00 | 30.00 | 0.30 | 30 |
| 31.00 | 40.00 | 0.40 | 40 |
+----------+---------+-----------+------+
The first problem is to work out how commission ranges map to sales totals to determine base amounts for calculation of each part-commission. We assume the ranges are inclusive, ie a range that starts at 1 euro is meant to include that first euro:
- if amt < comstart, base amount = 0
- if amt <= commend, base amount = amt-comstart+1
- if amt > commend, base amount = commend - comstart+1
This is a nested IF():
IF(s.amt<c.comstart,0,IF(s.amt<=c.commend,s.amt-c.comstart,c.commend-c.comstart))
The second problem is how to apply every commission range row to every employee sales sum. That's a
CROSS JOIN between aggregated sales and commissions:
SELECT *
FROM (
SELECT employeeID,SUM(sales) AS amt
FROM sales
GROUP BY employeeID
) AS s
JOIN commissions
ORDER BY s.employeeID;
+------------+------+----------+---------+-----------+------+
| employeeID | amt | comstart | commend | comfactor | pct |
+------------+------+----------+---------+-----------+------+
| 1 | 14 | 1.00 | 10.00 | 0.10 | 10 |
| 1 | 14 | 11.00 | 20.00 | 0.20 | 20 |
| 1 | 14 | 21.00 | 30.00 | 0.30 | 30 |
| 1 | 14 | 31.00 | 40.00 | 0.40 | 40 |
| 2 | 36 | 31.00 | 40.00 | 0.40 | 40 |
| 2 | 36 | 1.00 | 10.00 | 0.10 | 10 |
| 2 | 36 | 11.00 | 20.00 | 0.20 | 20 |
| 2 | 36 | 21.00 | 30.00 | 0.30 | 30 |
+------------+------+----------+---------+-----------+------+
Now check how the formula applies on every commission band for every sales total:
SELECT
s.employeeID,s.amt,c.comstart,c.commend,
IF(s.amt<=c.comstart,0,
IF( s.amt < c.commend, s.amt-c.comstart+1, c.commend-c.comstart+1 )
) AS base,
c.comFactor AS ComPct,
IF(s.amt<=c.comstart,0,
IF(s.amt<c.commend,s.amt-c.comstart+1,c.commend-c.comstart+1)
) * comFactor AS Comm
FROM (
SELECT employeeID,SUM(sales) AS amt
FROM sales
GROUP BY employeeID
) AS s
JOIN commissions c
ORDER BY s.employeeID,comstart;
+------------+------+----------+---------+-------+--------+--------+
| employeeID | amt | comstart | commend | base | ComPct | Comm |
+------------+------+----------+---------+-------+--------+--------+
| 1 | 14 | 1.00 | 10.00 | 10.00 | 0.10 | 1.0000 |
| 1 | 14 | 11.00 | 20.00 | 4.00 | 0.20 | 0.8000 |
| 1 | 14 | 21.00 | 30.00 | 0.00 | 0.30 | 0.0000 |
| 1 | 14 | 31.00 | 40.00 | 0.00 | 0.40 | 0.0000 |
| 2 | 36 | 1.00 | 10.00 | 10.00 | 0.10 | 1.0000 |
| 2 | 36 | 11.00 | 20.00 | 10.00 | 0.20 | 2.0000 |
| 2 | 36 | 21.00 | 30.00 | 10.00 | 0.30 | 3.0000 |
| 2 | 36 | 31.00 | 40.00 | 6.00 | 0.40 | 2.4000 |
+------------+------+----------+---------+-------+--------+--------+
Finally,
SUM formula results to aggregate commissions on aggregated sales:
SELECT
s.employeeID,
ROUND(s.amt,2) AS Amount,
ROUND( SUM(IF(s.amt<=c.comstart,0,
IF(s.amt<=c.commend,s.amt-c.comstart+1,c.commend-c.comstart+1)
) * c.pct/100),2 ) AS Commissions
FROM (
SELECT employeeID,SUM(sales) AS amt
FROM sales
GROUP BY employeeID
) AS s
JOIN commissions c
GROUP BY s.employeeID;
+------------+--------+-------------+
| employeeID | Amount | Commissions |
+------------+--------+-------------+
| 1 | 14.00 | 2.00 |
| 2 | 36.00 | 8.00 |
+------------+--------+-------------+
Here is another example. We track passenger flight bookings in three tables: flight, booking, passenger. To report all destinations per passenger, retrieve
DISTINCT passenger-destination combos, then count them:
DROP TABLES IF EXISTS flight, booking, passenger;
CREATE TABLE flight(flight CHAR(12),source CHAR(12),destination CHAR(12));
INSERT INTO flight VALUES
('ab123','dublin','london'),('bc123','prague','london'),('cd123','stuttgart','paris'),('de123','paris','madrid');
CREATE TABLE booking(flight CHAR(5),seat CHAR(2),passenger_id INT NOT NULL);
INSERT INTO booking VALUES
('ab123','a2',1),('bc123','a1',2),('bc123','a2',1),('cd123','a3',1);
CREATE TABLE passenger(passenger_id INT, name VARCHAR(12));
INSERT INTO passenger VALUES (1,'john'),(2,'bill'),(3,'david');
SELECT x.*,COUNT(b.passenger_id) bookings
FROM (
SELECT DISTINCT p.passenger_id, p.name, d.destination
FROM passenger p
CROSS JOIN flight d
) x
LEFT JOIN flight d ON d.destination = x.destination
LEFT JOIN booking b ON b.passenger_id = x.passenger_id AND b.flight = d.flight
GROUP BY passenger_id, destination;
+--------------+-------+-------------+----------+
| passenger_id | name | destination | bookings |
+--------------+-------+-------------+----------+
| 1 | john | london | 2 |
| 1 | john | madrid | 0 |
| 1 | john | paris | 1 |
| 2 | bill | london | 1 |
| 2 | bill | madrid | 0 |
| 2 | bill | paris | 0 |
| 3 | david | london | 0 |
| 3 | david | madrid | 0 |
| 3 | david | paris | 0 |
+--------------+-------+-------------+----------+
Show only one child row per parent row
Given tables
parent(id int not null primary key, etc...) and
child (id int not null primary key, pid int not null references parent (id), etc...), how do we write a query that retrieves only one child row per
pid even when the child table has multiple matching rows? MySQL permits use of
GROUP BY even when the
SELECT list specifies no aggregate function, so this will work:
select p.id, c.id
from parent p
join child c on p.id=c.pid
group by p.id;
But is it accurate? No, because it displays only the first c.pid value it happens to find. For further discussion see
Within-group aggregates.
Skip repeating values
You want to report all unique values of a column and skip all rows repeating any of these values.
SELECT col
FROM foo
GROUP BY col
Values linked with all values of another column
You have a table in which each row references one text and one keyword in the text ...
DROP TABLE IF EXISTS keywords;
CREATE TABLE keywords (txtID int, keyword char(8));
INSERT INTO keywords VALUES(1 , 'foo'),(2 , 'bar'),(1 , 'foo'),(2 , 'foo');
... and you want a list of texts which include every keyword.
You might think you have to join and match. You don't. All you need to do is count the distinct keywords which occur for each text, then for each text compare that number with the entire list of distinct keywords:
SELECT txtID, COUNT(DISTINCT keyword) AS N
FROM keywords
GROUP BY txtID
HAVING N = (SELECT COUNT(DISTINCT keyword) FROM keywords);
+-------+---+
| txtID | N |
+-------+---+
| 2 | 2 |
+-------+---+
Within-group aggregates
You have a
products table with columns
item, supplier, price. Multiple suppliers offer various prices for the same item. You need to find the supplier with the lowest price for each item.
DROP TABLE IF EXISTS products;
CREATE TABLE products(item int,supplier int,price decimal(6,2));
INSERT INTO products VALUES(1,1,10),(1,2,15),(2,2,20),(2,1,21),(2,2,18);
SELECT * FROM products;
+------+----------+-------+
| item | supplier | price |
+------+----------+-------+
| 1 | 1 | 10.00 |
| 1 | 2 | 15.00 |
| 2 | 2 | 20.00 |
| 2 | 1 | 21.00 |
| 2 | 2 | 18.00 |
+------+----------+-------+
Your first thought may be to
GROUP BY item, but that is not guaranteed to return the correct
supplier value for each minimum item price. Grouping by both
item and
supplier will return more information than you want. Nor can you write
WHERE price=MIN(...) because the query engine will evaluate the
WHERE clause before it knows the
MIN value.
This is the problem of
aggregating within aggregates. It is sometimes called the 'groupwise aggregates' problem, but the term 'groupwise' is ambiguous. We think better names for it are
subaggregates,
inner aggregates, or
within-group aggregates.
It's easy to show that the within-group aggregates problem is a form of the problem of returning values from non-grouping columns in an aggregate query. Suppose you write ...
SELECT item,supplier,MIN(price)
FROM products
GROUP BY item;
Will this reliably return the correct supplier per item? No.
Unless there is exactly one supplier per item, the supplier value returned will be arbitrary. To retrieve the correct supplier for each item, you need more logic.
One way to model the within-aggregates problem is via an
left self exclusion join...
SELECT p1.item,p1.supplier,p1.price
FROM products AS p1
LEFT JOIN products AS p2 ON p1.item = p2.item AND p1.price > p2.price
WHERE p2.item IS NULL;
...because in the resultset built by joining on left item = right item and left price > right price, the left-sided rows for which there is no greater right-sided price are precisely the per-item rows with the smallest prices.
Another solution is to derive an intermediate table of aggregated minimum prices. Before MySQL 4.1, it has to be a temporary table:
CREATE TEMPORARY TABLE tmp (
item INT,
minprice DECIMAL DEFAULT 0.0
);
LOCK TABLES products READ;
INSERT INTO tmp
SELECT item, MIN(price)
FROM products
GROUP BY item;
to which you then join the products table to find the matching suppliers:
SELECT products.item, supplier, products.price
FROM products
JOIN tmp ON products.item = tmp.item
WHERE products.price=tmp.minprice;
UNLOCK TABLES;
DROP TABLE tmp;
From MySQL 4.1 on, the temporary table can be
a correlated subquery. This is the most intuitively obvious syntax for the problem. Often it's also the
slowest solution:
SELECT item, supplier, price
FROM products AS p1
WHERE price = (
SELECT MIN(p2.price)
FROM products AS p2
WHERE p1.item = p2.item
);
Another solution, sometimes the fastest of all, is to move the aggregating subquery from the
WHERE clause to the
FROM clause:
SELECT p.item, p.supplier, p.price
FROM products AS p
JOIN (
SELECT item, MIN(price) AS minprice
FROM products
GROUP BY item
) AS pm ON p.item = pm.item AND p.price = pm.minprice;
Some users have trouble mapping elements of this model to their instance of the problem. There are five elements (or sets of them):
(i) a table, which might be a view, a single physical table, or a table derived from joins
(ii) one or more grouping columns,
(iii) one or more columns to aggregate,
(iv) one or more columns not mentioned in the
GROUP BY clause,
(v) an aggregating job to do, typically MIN() or MAX().
In the product/minimum price solution above:
(i) table
tbl = product
(ii) grouping column
grouping_col = item
(iii) column to aggregate =
col_to_aggregate = price
(iv) non-aggregated columns
other_detail, ...etc... = supplier
(v) aggregating function =
MIN().
Finally, here is a radically different model of the problem. It can find both within-group minima and within-group maxima in a single query. This model aggregates the concatenated within-group grouped column value and the within-group grouping column name in a single string, then uses
SUBSTR() to break them apart in the result:
SELECT
item,
SUBSTR( MIN( CONCAT( LPAD(price,6,0),supplier) ), 7) AS MinSupplier,
LEFT( MIN( CONCAT( LPAD(price,6,0),supplier) ), 6)+0 AS MinPrice,
SUBSTR( MAX( CONCAT( LPAD(price,6,0),supplier) ), 7) AS MaxSupplier,
LEFT( MAX( CONCAT( LPAD(price,6,0),supplier) ), 6)+0 AS MaxPrice
FROM products
GROUP BY item;
+------+-------------+----------+-------------+----------+
| item | MinSupplier | MinPrice | MaxSupplier | MaxPrice |
+------+-------------+----------+-------------+----------+
| 1 | 1 | 10 | 2 | 15 |
| 2 | 2 | 18 | 1 | 21 |
+------+-------------+----------+-------------+----------+
Try all solutions to find which is fastest for your version of the problem.
To find the top or bottom N per group, you might think the
LIMIT clause would work, but
LIMIT is limited in subqueries. See
Within-group quotas.
Within-group aggregates with a wrinkle
We have a wages table holding wage rates by waiter and startdate, and a tips table which tracks hours worked and tips received per waiter per day. The requirement is to report wages and concurrent tips per waiter per day.
DROP TABLE IF EXISTS wages,tips;
CREATE TABLE wages( id int, waiter int, start date, rate decimal(6,2));
INSERT INTO wages VALUES
( 1, 4, '2005-01-01', 5.00 ),
( 2, 4, '2005-03-01', 6.00 ),
( 3, 5, '2007-01-05', 7.00 ),
( 4, 5, '2008-03-20', 8.00 ),
( 5, 5, '2008-04-01', 9.00 );
CREATE TABLE tips(
id int,
date date,
waiter int,
hours_worked smallint,
tabs smallint,
tips decimal(6,2)
);
INSERT INTO tips VALUES
( 1, '2008-02-29', 4, 6.50, 21, 65.25 ),
( 2, '2008-03-06', 5, 6.00, 15, 51.75 ),
( 3, '2008-03-21', 4, 2.50, 5, 17.85 ),
( 4, '2008-03-22', 5, 5.25, 10, 39.00 );
SELECT * FROM wages;
+------+--------+------------+------+
| id | waiter | start | rate |
+------+--------+------------+------+
| 1 | 4 | 2005-01-01 | 5.00 |
| 2 | 4 | 2005-03-01 | 6.00 |
| 3 | 5 | 2007-01-05 | 7.00 |
| 4 | 5 | 2008-03-20 | 8.00 |
| 5 | 5 | 2008-04-01 | 9.00 |
+------+--------+------------+------+
SELECT * FROM tips;
+------+------------+--------+--------------+------+-------+
| id | date | waiter | hours_worked | tabs | tips |
+------+------------+--------+--------------+------+-------+
| 1 | 2008-02-29 | 4 | 7 | 21 | 65.25 |
| 2 | 2008-03-06 | 5 | 6 | 15 | 51.75 |
| 3 | 2008-03-21 | 4 | 3 | 5 | 17.85 |
| 4 | 2008-03-22 | 5 | 5 | 10 | 39.00 |
+------+------------+--------+--------------+------+-------+
For the above dataset, the result which correctly matches wages and tips would be:
+------+------------+------+------+-------+------+--------+------+------------+
| tid | Date | Hrs | tabs | tips | wid | waiter | rate | start |
+------+------------+------+------+-------+------+--------+------+------------+
| 1 | 2008-02-29 | 7 | 21 | 65.25 | 2 | 4 | 6.00 | 2005-03-01 |
| 2 | 2008-03-06 | 6 | 15 | 51.75 | 3 | 5 | 7.00 | 2007-01-05 |
| 3 | 2008-03-21 | 3 | 5 | 17.85 | 2 | 4 | 6.00 | 2005-03-01 |
| 4 | 2008-03-22 | 5 | 10 | 39.00 | 4 | 5 | 8.00 | 2008-03-20 |
+------+------------+------+------+-------+------+--------+------+------------+
Why is this different from an ordinary within-groups aggregate? The correct wage rate for a tips row is not the wages row for that waiter with the latest date; it is the wages row having the latest date before the date in the given tips row.
One way to proceed is to build a temporary table from a join of wages to tips on waiter and wages.start < tips.date, then exclusion-join that result to itself to remove all rows except those with the latest wage rate per tips row. A two-step:
-- wages-tips join removing later wage changes:
DROP TABLE IF EXISTS tmp;
CREATE TABLE tmp
SELECT
t.id AS tid, t.date AS Date, t.hours_worked AS Hrs,t.tabs,t.tips,
w.id AS wid, w.waiter, w.rate, w.start
FROM tips t
JOIN wages w ON w.waiter=t.waiter AND w.start<=t.date;
-- self-exclusion join to remove obsolete wage rows:
SELECT t1.*
FROM tmp t1
LEFT JOIN tmp t2 ON t1.tid=t2.tid and t1.start<t2.start
WHERE t2.waiter is null
ORDER BY t1.Date;
+------+------------+------+------+-------+------+--------+------+------------+
| tid | Date | Hrs | tabs | tips | wid | waiter | rate | start |
+------+------------+------+------+-------+------+--------+------+------------+
| 1 | 2008-02-29 | 7 | 21 | 65.25 | 2 | 4 | 6.00 | 2005-03-01 |
| 2 | 2008-03-06 | 6 | 15 | 51.75 | 3 | 5 | 7.00 | 2007-01-05 |
| 3 | 2008-03-21 | 3 | 5 | 17.85 | 2 | 4 | 6.00 | 2005-03-01 |
| 4 | 2008-03-22 | 5 | 10 | 39.00 | 4 | 5 | 8.00 | 2008-03-20 |
+------+------------+------+------+-------+------+--------+------+------------+
DROP TABLE tmp;
That's fine, but can we skip the temp table? Yes—by adding the condition wages.start <= tips.date to each side of the exclusion join:
SELECT
t.id AS tid, t.date, t.hours_worked AS Hrs,t.tabs,t.tips,
w.id AS wid, w.waiter, w.rate, w.start
FROM tips t
JOIN wages w ON w.waiter=t.waiter AND w.start <= t.date
LEFT JOIN wages w2 ON w.waiter=w2.waiter AND w2.start<=t.date AND w.start<w2.start
WHERE w2.id IS NULL
ORDER BY t.date;
Much simpler, and it gives the same result as the two-step.
Within-group quotas (Top N per group)
A table has multiple rows per key value, and you need to retrieve, say, the first or earliest two rows per key.
If the groups are fairly small, this can be done efficiently with a self-join and counts. For example the following table (based on a tip by
Rudy Limeback) has three small data groups:
DROP TABLE IF EXISTS test;
CREATE TABLE test (
id INT,
entrydate DATE
);
INSERT INTO test VALUES
( 1, '2007-5-01' ),
( 1, '2007-5-02' ),
( 1, '2007-5-03' ),
( 1, '2007-5-04' ),
( 1, '2007-5-05' ),
( 1, '2007-5-06' ),
( 2, '2007-6-01' ),
( 2, '2007-6-02' ),
( 2, '2007-6-03' ),
( 2, '2007-6-04' ),
( 3, '2007-7-01' ),
( 3, '2007-7-02' ),
( 3, '2007-7-03' );
The first two rows per
ID are the rows which, for a given
ID, have two or fewer rows with earlier dates. If we use an
inequality join with the
COUNT(*) function to find the earlier rows per
ID ...
SELECT t1.id, t1.entrydate, COUNT(*) AS earlier
FROM test AS t1
JOIN test AS t2 ON t1.id=t2.id AND t1.entrydate >= t2.entrydate
GROUP BY t1.id, t1.entrydate
+------+------------+---------+
| id | entrydate | earlier |
+------+------------+---------+
| 1 | 2007-05-01 | 1 |
| 1 | 2007-05-02 | 2 |
| 1 | 2007-05-03 | 3 |
| 1 | 2007-05-04 | 4 |
| 1 | 2007-05-05 | 5 |
| 1 | 2007-05-06 | 6 |
| 2 | 2007-06-01 | 1 |
| 2 | 2007-06-02 | 2 |
| 2 | 2007-06-03 | 3 |
| 2 | 2007-06-04 | 4 |
| 3 | 2007-07-01 | 1 |
| 3 | 2007-07-02 | 2 |
| 3 | 2007-07-03 | 3 |
+------+------------+---------+
... then we get our result immediately by removing rows where the 'earlier' count exceeds 2:
SELECT t1.id, t1.entrydate, count(*) AS earlier
FROM test AS t1
JOIN test AS t2 ON t1.id=t2.id AND t1.entrydate >= t2.entrydate
GROUP BY t1.id, t1.entrydate
HAVING earlier <= 2;
+------+------------+---------+
| id | entrydate | earlier |
+------+------------+---------+
| 1 | 2007-05-01 | 1 |
| 1 | 2007-05-02 | 2 |
| 2 | 2007-06-01 | 1 |
| 2 | 2007-06-02 | 2 |
| 3 | 2007-07-01 | 1 |
| 3 | 2007-07-02 | 2 |
+------+------------+---------+
This works beautifully with smallish aggregates. But the query algorithm compares every within-group row to every other within-group row. As the size N of a group increases, execution time increases by N*N. If the query takes one minute for groups of 1,000, it will take 16 minutes for groups of 4,000, and more than four hours for groups for 16,000.
The solution does not scale.
What to do? Forget
GROUP BY! Manually assemble the desired query results in a temporary table from simple indexed queries, in this case, two rows per
ID:
DROP TEMPORARY TABLE IF EXISTS earliers;
CREATE TEMPORARY TABLE earliers( id INT, entrydate DATE);
INSERT INTO earliers
SELECT id,entrydate FROM test WHERE id=1 ORDER BY entrydate LIMIT 2;
INSERT INTO earliers
SELECT id,entrydate FROM test WHERE id=2 ORDER BY entrydate LIMIT 2;
INSERT INTO earliers
SELECT id,entrydate FROM test WHERE id=3 ORDER BY entrydate LIMIT 2;
You need one
INSERT statement per grouping value. To print the result, just query the
earliers table:
SELECT * FROM earliers
ORDER BY id, entrydate;
+------+------------+
| id | entrydate |
+------+------------+
| 1 | 2007-05-01 |
| 1 | 2007-05-02 |
| 2 | 2007-06-01 |
| 2 | 2007-06-02 |
| 3 | 2007-07-01 |
| 3 | 2007-07-02 |
+------+------------+
DROP TEMPORARY TABLE earliers;
Most useful reports run again and again. If that's the case for yours, automate it in a stored procedure: using a cursor and a prepared statement, auto-generate an
INSERT statement for every grouping value, and return the result:
DROP PROCEDURE IF EXISTS listearliers;
DELIMITER |
CREATE PROCEDURE listearliers()
BEGIN
DECLARE curdone, vid INT DEFAULT 0;
DECLARE idcur CURSOR FOR SELECT DISTINCT id FROM test;
DECLARE CONTINUE HANDLER FOR SQLSTATE '02000' SET curdone = 1;
DROP TEMPORARY TABLE IF EXISTS earliers;
CREATE TEMPORARY TABLE earliers( id INT, entrydate DATE);
SET @sql = 'INSERT INTO earliers SELECT id,entrydate FROM test WHERE id=? ORDER BY entrydate LIMIT 2';
OPEN idcur;
REPEAT
FETCH idcur INTO vid;
IF NOT curdone THEN
BEGIN
SET @vid = vid;
PREPARE stmt FROM @sql;
EXECUTE stmt USING @vid;
DROP PREPARE stmt;
END;
END IF;
UNTIL curdone END REPEAT;
CLOSE idcur;
SELECT * FROM earliers ORDER BY id,entrydate;
DROP TEMPORARY TABLE earliers;
END;
|
DELIMITER ;
CALL listearliers();
Aggregates from bands of values
To count and average scores in bands of 10, ie 0-9,10-19 etc:
create table scores(score int);
insert into scores values(5),(15),(25),(35);
SELECT 10 * FLOOR( score / 10 ) AS Bottom,
10 * FLOOR( score / 10 ) + 9 AS Top,
Count( score ),
Avg( score )
FROM scores
GROUP BY 10 * FLOOR( score / 10 );
+--------+------+----------------+--------------+
| Bottom | Top | Count( score ) | Avg( score ) |
+--------+------+----------------+--------------+
| 0 | 9 | 1 | 5.0000 |
| 10 | 19 | 1 | 15.0000 |
| 20 | 29 | 1 | 25.0000 |
| 30 | 39 | 1 | 35.0000 |
+--------+------+----------------+--------------+
Here is an example from a MySQL forum. You have a table of IDs and point scores, and you wish to show the distribution of counts across ranges. A traditional way of doing this is to build a table of ranges, then join from ranges to scores:
drop table if exists points,ranges;
create table points (
id int not null auto_increment primary key,
points int not null
);
create table ranges (
low int not null,
high int not null
);
insert into points (points) values
(456),(401),(543),(234),(303),(521),(478),(643),(575),(456),(432),
(312),(564),(423),(411),(395),(543);
insert into ranges (low, high) values
(0,50),(51,100),(101,150),(151,200),(201,250),(251,300),(301,350),
(351,400),(401,450),(451,500),(501,550),(551,600),(601,650),(651,700);
select r.low, r.high, count(p.id)
from ranges r
left join points p on p.points between r.low and r.high
group by r.low, r.high;
That's fine, but the ranges should be consistent, eg 0-49,50-99,100-149..., and if you keep a simple utility table of ints 0..9, you never need to build another range table:
drop table if exists ints,ranges;
create table ints(i tinyint);
insert into ints values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
select ranges.low, ranges.high, count(p.id)
from (
select ( t.i*100 + u.i*10 + v.i ) as low,( t.i*100 + u.i*10 + v.i ) + 49 as high
from ints t
join ints u
join ints v
where (t.i*100 + u.i*10 + v.i) mod 50 = 0 and (t.i*100 + u.i*10 + v.i) < 701
) ranges
left join points p on p.points between ranges.low and ranges.high
group by ranges.low;
+-----+------+-------------+
| low | high | count(p.id) |
+-----+------+-------------+
| 0 | 50 | 0 |
| 51 | 100 | 0 |
| 101 | 150 | 0 |
| 151 | 200 | 0 |
| 201 | 250 | 1 |
| 251 | 300 | 0 |
| 301 | 350 | 2 |
| 351 | 400 | 1 |
| 401 | 450 | 4 |
| 451 | 500 | 3 |
| 501 | 550 | 3 |
| 551 | 600 | 2 |
| 601 | 650 | 1 |
| 651 | 700 | 0 |
+-----+------+-------------+
Last updated 15 Oct 2009
Average the top 50% values per group
Each row of a games table records one game score for a team:
DROP TABLE IF EXISTS games;
CREATE TABLE games(id INT, teamID INT, score INT);
INSERT INTO games VALUES
(1,1,3),(2,1,4),(3,1,5),(4,1,6),(5,2,6),
(6,2,7),(7,2,8),(8,2,7),(9,2,6),(10,2,7);
How would we write a query that returns the average of the top 50% of scores per team?
The per-team median value is its middle value--lower than the highest 50% and higher than the lowest 50% of values for that team--so a shortcut is to query the team medians, then aggregate on a join that selects per-team scores above the medians.
How to find per-team medians? If a resultset has an odd number of rows, at least one row has the true median score. If it has an even number of rows, the median score is an average of two central values. The following query adapts Joe Celko's formula in "SQL for Smarties" averaging "low" and "high" medians:
DROP TABLE IF EXISTS medians;
CREATE TABLE medians
SELECT p1.teamid, AVG(P1.score) AS median
FROM games AS P1, games AS P2
WHERE p1.teamid=p2.teamid
GROUP BY p1.teamid
HAVING (
SUM(CASE WHEN P2.score <= P1.score THEN 1 ELSE 0 END) >= ((COUNT(*) + 1) / 2)
AND
SUM(CASE WHEN P2.score >= P1.score THEN 1 ELSE 0 END) >= (COUNT(*)/2 + 1)
)
OR (
SUM(CASE WHEN P2.score >= P1.score THEN 1 ELSE 0 END) >= ((COUNT(*) + 1) / 2)
AND
SUM(CASE WHEN P2.score <= P1.score THEN 1 ELSE 0 END) >= (COUNT(*)/2 + 1)
);
+--------+--------+
| teamid | median |
+--------+--------+
| 1 | 4.5000 |
| 2 | 6.8333 |
+--------+--------+
Now join games to medians accepting only top-half values:
SELECT g.teamid, AVG(g.score) AS Top50Avg
FROM games g
JOIN medians m ON g.teamid = m.teamid AND g.score >= m.median
GROUP BY g.teamid
ORDER BY Top50Avg DESC;
+--------+----------+
| teamid | Top50Avg |
+--------+----------+
| 2 | 7.2500 |
| 1 | 5.5000 |
+--------+----------+
DROP TABLE medians;
Yes, all the logic can be moved into one query:
SELECT g.teamid, AVG(g.score) AS Top50Avg
FROM games g
JOIN (
SELECT p1.teamid, AVG(P1.score) AS median
FROM games AS P1, games AS P2
WHERE p1.teamid=p2.teamid
GROUP BY p1.teamid
HAVING (
SUM(CASE WHEN P2.score <= P1.score THEN 1 ELSE 0 END) >= ((COUNT(*) + 1) / 2)
AND
SUM(CASE WHEN P2.score >= P1.score THEN 1 ELSE 0 END) >= (COUNT(*)/2 + 1)
)
OR (
SUM(CASE WHEN P2.score >= P1.score THEN 1 ELSE 0 END) >= ((COUNT(*) + 1) / 2)
AND
SUM(CASE WHEN P2.score <= P1.score THEN 1 ELSE 0 END) >= (COUNT(*)/2 + 1)
)
) AS m ON g.teamid = m.teamid AND g.score >= m.median
GROUP BY g.teamid
ORDER BY Top50Avg DESC;
Correlation
Probably more bunk has been written about correlation than about any other statistic. We'll keep this short and straight. At its simplest, correlation is a statistical measure of non-random, linear association between pairs of values in a dataset. It's denoted by r, and varies from -1 through +1, where -1 indicates perfect inverse correlation (the regression line goes down left to right), 0 indicates no correlation (there is no regression line; it's just a scatterplot), and +1 indicates perfect direct correlation (the regression line goes up left to right).
For an example we'll use a bit of imaginary data:
drop table if exists t;
create table t (id int, x int, y float);
insert into t values
(1 , 68, 4.1),(2 , 71, 4.6),(3 , 62, 3.8),(4 , 75, 4.4),(5 , 58, 3.2),
(6 , 60, 3.1),(7 , 67, 3.8),(8 , 68, 4.1),(9 , 71, 4.3),(10, 69, 3.7),
(11, 68, 3.5),(12, 67, 3.2),(13, 63, 3.7),(14, 62, 3.3),(15, 60, 3.4),
(16, 63, 4.0),(17, 65, 4.1),(18, 67, 3.8),(19, 63, 3.4),(20, 61, 3.6);
If you like to think about such problems concretely, you can think of id as a subject's id, x as a subject's height, and y as a subject's score on a self-confidence questionnaire, so we would be computing a correlation between height and self-confidence.
There are many correlation formulas. Most commonly used is the Pearson product-moment correlation coefficient, which is valid only for normally distributed data (data which roughly fits a bell curve). A good Wikipedia page offers several formulas for the Pearson coefficient. Note that when the data skews significantly from a normal distribution, you very likely need a different formula.
That's one problem with correlation: it isn't always a valid number to calculate. Another problem is that computing just the correlation coefficient is never enough. You also need to find the probability that the coefficient you calculated is significantly different from 0; that is something you look up in a statistical table, and it's outside our scope here. You will probably also want the slope of the relationship or regression line, its intercept with the Y axis, and the coefficient of regression, which is the proportion of observed variation due to correlation.
For this example we assume a normal distribution and no missing values.
Calculation of the correlation coefficient needs two passes: a first to calculate basic statistical quantities, then a second to calculate the slope, intercept and correlation coefficient from those basic quantities. Calculating the coefficient of regression requires a third pass.
If we're computing correlation in a database like MySQL, then, you'd expect Views will be useful. Unfortunately we're immediately bitten by two limitations in the MySQL implementation of Views: subqueries in the FROM clause of a View are not supported, and neither are parameters.
Then do we need to encapsulate correlation in a stored procedure? We could, but we needn't because of how r calculates. We like a more direct solution. The algorithm is:
1. Calculate the required basic statistics.
2. Use the results of #1 to calculate slope, intercept and r.
3. Use the slope and intercept from #2 to calculate the coefficient of regression.
4. Collect and display the results.
In this solution we do it all in one query ...
1. The innermost query does step 1,
2. The query that references the inner query does step 2,
3. The next outer query calculates the regression coefficient by cross-joining the step 2 query result, which is one logical row, with the original table.
4. The outermost query displays the result:
Read the query inside out:
SELECT -- Step 4
N, Slope, avgY - slope*avgX AS Intercept,
Correlation, CoeffOfReg
FROM (
SELECT -- Step 3
N, avgX, avgY, slope, intercept, Correlation,
FORMAT( 1 - SUM((y - intercept - slope*x)*(y - intercept - slope*x))/
((N-1)*varY), 5 ) AS CoeffOfReg
FROM t AS t2
JOIN (
SELECT -- Step 2
N, avgX, avgY, varY, slope,
Correlation, avgY - slope*avgX AS intercept
FROM (
SELECT
N, avgX, avgY, varY,
FORMAT(( N*sumXY - sumX*sumY ) /
( N*sumsqX - sumX*sumX ), 5 ) AS slope,
FORMAT(( sumXY - n*avgX*avgY ) /
( (N-1) * SQRT(varX) * SQRT(varY)), 5 ) AS Correlation
FROM (
SELECT -- Step 1
COUNT(x) AS N,
AVG(x) AS avgX,
SUM(x) AS sumX,
SUM(x*x) AS sumsqX,
VAR_SAMP(x) AS varX,
AVG(y) AS avgY,
SUM(y) AS sumY,
SUM(y*y) AS sumsqY,
VAR_SAMP(y) AS varY,
SUM(x*y) AS sumXY
FROM t
) AS sums
) AS calc
) stats
) combined;
+----+---------+---------------------+-------------+------------+
| N | Slope | Intercept | Correlation | CoeffOfReg |
+----+---------+---------------------+-------------+------------+
| 20 | 0.07066 | -0.8661640047683719 | 0.73064 | 0.53383 |
+----+---------+---------------------+-------------+------------+
Last updated 22 May 2009
Count unique values of one column
SELECT col_name, COUNT(*) AS frequency
FROM tbl_name
GROUP by col_name
ORDER BY frequency DESC;
Median
Statistically, the median is the middle value--the value that is smaller than that found in half of all remaining rows, and larger than that found in the other half:
SELECT l1.hours As Median
FROM BulbLife As l1, bulbLife AS l2
GROUP BY l1.Hours
HAVING SUM(CASE WHEN l2.hours <= l1.hours THEN 1 ELSE 0 END) >= (COUNT(*)+1) / 2
AND SUM(CASE WHEN l2.hours >= l1.hours THEN 1 ELSE 0 END) >= (COUNT(*)/2) + 1;
An anonymous reader pointed out that this will cost O(N*N), ie it does not scale, so we posted a MySQL implementation of Torben Mogenson's algorithm for calculating the median (
http://ndevilla.free.fr/median/median/node20.html), which is said to be amongst the fastest. It also proved too slow. Now Joe Wynne has offered an algorithm which appears to be correct, and which does scale. Here it is as a MySQL stored procedure:
DROP PROCEDURE IF EXISTS Median;
DELIMITER |
CREATE PROCEDURE Median( tbl CHAR(64), col CHAR(64), OUT res DOUBLE )
BEGIN
DECLARE arg CHAR(64);
SET @sql = CONCAT( 'SELECT ((COUNT(*))/2) INTO @c FROM ', tbl );
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
SET @a = CONVERT(FLOOR(@c), SIGNED);
IF @a = @c THEN
BEGIN
SET @a = @a-1;
SET @b = 2;
SET arg = CONCAT( 'AVG(', col, ')' );
END;
ELSE
BEGIN
SET @b = 1;
SET arg = col;
END;
END IF;
SET @sql = CONCAT('SELECT ', arg, ' INTO @res FROM (SELECT ', col, ' FROM ', tbl,
' ORDER BY ', col, ' LIMIT ?,?) as tmp');
PREPARE stmt FROM @sql;
EXECUTE stmt USING @a, @b;
DROP PREPARE stmt;
SET res=@res;
END;
|
DELIMITER ;
Why don't we make it a function? Because MySQL functions do not (yet?) allow dynamic SQL.
Mode
Statistically, the mode is the most frequently occurring value. Given tables parent(id int) and child(pid int, cid int), where child.pid references parent.id as a foreign key, what query finds the parent.id most often represented in the child id, that is, the modal count of child.pid?
SELECT pid, COUNT(*) AS frequency
FROM child
GROUP BY pid
ORDER BY frequency DESC
LIMIT 1;
Rank order
Without MSSQL's
RANK() aggregate function, how do we display rank order in a MySQL query, for example from a table like this?
CREATE TABLE votes( name CHAR(10), votes INT );
INSERT INTO votes VALUES
('Smith',10),('Jones',15),('White',20),('Black',40),('Green',50),('Brown',20);
The query is a two-step:
1. Join the table to itself on the value to be ranked, handling ties
2. Group and order the result of the self-join on rank:
SELECT v1.name, v1.votes, COUNT(v2.votes) AS Rank
FROM votes v1
JOIN votes v2 ON v1.votes < v2.votes OR (v1.votes=v2.votes and v1.name = v2.name)
GROUP BY v1.name, v1.votes
ORDER BY v1.votes DESC, v1.name DESC;
+-------+-------+------+
| name | votes | Rank |
+-------+-------+------+
| Green | 50 | 1 |
| Black | 40 | 2 |
| White | 20 | 3 |
| Brown | 20 | 3 |
| Jones | 15 | 5 |
| Smith | 10 | 6 |
+-------+-------+------+
Suppose you want to know the vote count for White and the names and tallies for the next highest and next lowest counts:
SELECT
s1.name, s1.votes, COUNT(s2.name) rank,
IF(s1.name = 'White','<-',' ') AS 'Near Ranks'
FROM votes s1
JOIN votes s2 ON s1.votes < s2.votes or (s1.votes=s2.votes and s1.name = s2.name)
GROUP BY s1.name, s1.votes
ORDER BY rank
) a
LEFT JOIN (
SELECT
s1.name, s1.votes, COUNT(s2.name) rank,
IF(s1.name = 'White', '<-',' ') AS 'Near Ranks'
FROM votes s1
JOIN votes s2 ON s1.votes < s2.votes or (s1.votes=s2.votes and s1.name = s2.name)
GROUP BY s1.name, s1.votes
ORDER BY rank
) b ON a.rank BETWEEN b.rank-1 AND b.rank+1
WHERE a.name = 'White';
+-------+-------+------+------------+
| name | votes | rank | Near Ranks |
+-------+-------+------+------------+
| Black | 40 | 2 | |
| White | 20 | 3 | <- |
| Brown | 20 | 3 | |
+-------+-------+------+------------+
But these methods don't scale; they're O(N
2) or worse, especially if non-trivial calculation is needed to compute rank. If you do not need to identify ties in the ranking column, writing a temporary ordering table and then querying it is often much faster:
DROP TEMPORARY TABLE IF EXISTS tmp;
SET @i=0;
CREATE TEMPORARY TABLE tmp
SELECT idcol,valuecol,@i:=@i+1 AS rank
ORDER BY valuecol DESC;
Baron Schwartz has a
useful page on optimising rankings for leaderboards. It requires maintenance of a ranking computation table, as described on his page under "How to maintain the rank column". That table has this structure:
CREATE TABLE score_ranked (
gamer INT NOT NULL,
game INT NOT NULL,
score INT) NOT NULL,
rank_in_game INT NOT NULL DEFAULT 0,
PRIMARY KEY (gamer,game),
KEY (game,score),
KEY (score),
KEY (game,rank_in_game)
) ENGINE=InnoDB;
If you use Baron's schema, you will find the following query, developed on a MySQL forum by Rick James, reasonably efficient for retrieving top score sums for all games:
SELECT
@seq := 0, -- raw 1,2,3,...
@rank := 0, -- allow for ties
@prev := 999999999; -- catch ties
SELECT Rank, gamer, point
FROM (
SELECT
@seq := IF(@prev = point, @seq, @seq + 1) AS seq,
@rank := IF(@prev = point, @rank, @seq) AS Rank,
@prev := point as prev, # I forgot this
gamer,
point
FROM (
SELECT gamer, SUM(score) as point
FROM score_ranked s
GROUP BY gamer
ORDER BY point DESC
) x
) y
WHERE Rank <= 10
ORDER BY Rank, gamer;
Backslashes in data
Backslashes multiply weirdly:
SELECT 'a\b' RLIKE 'a\b';
returns 1, as does...
SELECT 'a\\b' RLIKE 'a\\\\b';
because in a pair of backslashes, the second is not escaped by the first, so to compare two literals you double each backslash in the RLIKE argument. But if you are querying a table for such a string from the MySQL client, this doubling happens twice--once in the client, and once in the database--so to find a column value matching 'a\\b', you need to write...
SELECT desc FROM xxx WHERE desc RLIKE 'aa\\\\\\\\bb';
That's eight backslashes to match two!
Compare data in two tables
This query UNIONs queries for matching column names from two tables, and keeps just those rows which occur once in the union. Those are the rows unique to one table or the other. Customise your column list { id, col1, col2, col3 ...} as desired.
SELECT
MIN(TableName) as TableName, id, col1, col2, col3, ...
FROM (
SELECT 'Table a' as TableName, a.id, a.col1, a.col2, a.col3, ...
FROM a
UNION ALL
SELECT 'Table b' as TableName, b.id, b.col1, b.col2, b.col3, ...
FROM b
) AS tmp
GROUP BY id, col1, col2, col3, ...
HAVING COUNT(*) = 1
ORDER BY ID;
Show rows where column value changed
SQL is set-oriented, but it can solve row-by-row problems. Suppose you need to retrieve only the rows that differ from immediatel previous rows given some ordering spec:
drop table if exists t;
create table t (
p char(3),
d date
);
insert into t values
('50%','2008-05-01'),
('30%','2008-05-02'),
('30%','2008-05-03'),
('50%','2008-05-04'),
('50%','2008-05-05'),
('20%','2008-05-06'),
('20%','2008-05-07'),
('50%','2008-05-08'),
('70%','2008-05-09'),
('70%','2008-05-10');
select * from t order by d;
+------+------------+
| p | d |
+------+------------+
| 50% | 2008-05-01 | *
| 30% | 2008-05-02 | *
| 30% | 2008-05-03 |
| 50% | 2008-05-04 | *
| 50% | 2008-05-05 |
| 20% | 2008-05-06 | *
| 20% | 2008-05-07 |
| 50% | 2008-05-08 | *
| 70% | 2008-05-09 | *
| 70% | 2008-05-10 |
+------+------------+
We want to retrieve only rows whose `p` values differ from immediately previous values (marked by * above). As with running sums. we get the desired listing by tracking row-to-row value changes with user variables:
set @p='';
set @d='';
select p 'Pct Changed',d Date from (
select
p,
if( p<>@p, d, @d ) as d,
@p:=p,
@d:=d
from t
order by d
) as t
group by d;
+-------------+------------+
| Pct Changed | Date |
+-------------+------------+
| 50% | 2008-05-01 |
| 30% | 2008-05-02 |
| 50% | 2008-05-04 |
| 20% | 2008-05-06 |
| 50% | 2008-05-08 |
| 70% | 2008-05-09 |
+-------------+------------+
Add auto-incrementing primary key to a table
The steps are: (i) recreate the table, populating a new column from an incrementing user variable, then (ii) alter the table to add auto_increment and primary key properties to the new column. So given table t with columns named `dt` and `observed`...
DROP TABLE IF EXISTS t2;
SET @id=0;
CREATE TABLE t2
SELECT @id:=@id+1 AS id, dt, observed FROM t ORDER BY dt;
ALTER TABLE t2
MODIFY id INT AUTO_INCREMENT PRIMARY KEY;
DROP TABLE t;
RENAME TABLE t2 TO t;
Auto-increment: reset next value
ALTER TABLE tbl SET AUTO_INCREMENT=val;
Change or drop a foreign key
To change a foreign key, first drop it, then declare the new, revised foreign key. The syntax for declaring a foreign key is ...
[CONSTRAINT [constraint_name]]
FOREIGN KEY [key_name] (keycol_name,...) reference_definition
and the syntax for dropping one is ...
DROP FOREIGN KEY constraint_name
Notice that you can omit the
CONSTRAINT when you declare a foreign key, but the
only way to
DROP a foreign key is to reference it by the
constraint_name which you probably never specified!
There should be a circle of hell reserved for designers who build inconsistencies like this into their tools. The only way round this one is to run
SHOW CREATE TABLE to find out what the foreign key's
constraint_name is, so you can write the
DROP statement. Here is a wee test case:
drop table if exists a,b;
create table a(i int primary key)engine=innodb;
create table b(i int,foreign key(i) references a(i)) engine=innodb;
show create table\G
CREATE TABLE `b` (
`i` int(11) DEFAULT NULL,
KEY `i` (`i`),
CONSTRAINT `b_ibfk_1` FOREIGN KEY (`i`) REFERENCES `a` (`i`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
-- drop and recreate the FK:
alter table b drop foreign key b_ibfk_1;
alter table b add foreign key(i) references a(i) on update cascade;
show create table b\G
Create Table: CREATE TABLE `b` (
`i` int(11) DEFAULT NULL,
KEY `i` (`i`),
CONSTRAINT `b_ibfk_1` FOREIGN KEY (`i`) REFERENCES `a` (`i`) ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
drop table a,b;
Compare structures of two tables
To compare columns by name and ordinal position in tables test.t1 and test.t2:
SELECT
MIN(TableName) AS 'Table',
column_name AS 'Column',
ordinal_position AS 'Position'
FROM (
SELECT
't1' as TableName,
column_name,
ordinal_position
FROM information_schema.columns AS i1
WHERE table_schema='test' AND table_name='t1'
UNION ALL
SELECT
't2' as TableName,
column_name,
ordinal_position
FROM information_schema.columns AS i2
WHERE table_schema='test' AND table_name='t2'
) AS tmp
GROUP BY column_name
HAVING COUNT(*) = 1
ORDER BY ordinal_position;
For MySQL 5.0.2 or later here is a query that lists all table structure differences between any two tables. It selects all information_schema.columns rows for one table, does the same for the second table, UNIONs these two queries, then uses HAVING to pick only those rows where the COUNT(*) in the union is 1—that is, where any column of one table differs from its mate.
To avoid having to cut and paste database and table names, save it as a stored procedure in any database (other than information_schema):
DROP PROCEDURE IF EXISTS CompareTableStructs;
-- Uncomment if MySQL version is 5.0.6-5.0.15:
-- SET GLOBAL log_bin_trust_routine_creators=TRUE;
DELIMITER |
CREATE PROCEDURE CompareTableStructs (
IN db1 CHAR(64), IN tbl1 CHAR(64), IN db2 CHAR(64), IN tbl2 CHAR(64)
)
SELECT
MIN(TableName) AS TableName,
column_name,
ordinal_position,
column_default,
is_nullable,
data_type,
character_maximum_length,
numeric_precision,
numeric_scale,
character_set_name,
collation_name,
column_type,
column_key,
extra,
privileges,
column_comment
FROM (
SELECT
tbl1 as TableName,
column_name,
ordinal_position,
column_default,
is_nullable,
data_type,
character_maximum_length,
numeric_precision,
numeric_scale,
character_set_name,
collation_name,
column_type,
column_key,
extra,
privileges,
column_comment
FROM information_schema.columns AS i1
WHERE table_schema=db1 AND table_name=tbl1
UNION ALL
SELECT
tbl2 as TableName,
column_name,
ordinal_position,
column_default,
is_nullable,
data_type,
character_maximum_length,
numeric_precision,
numeric_scale,
character_set_name,
collation_name,
column_type,
column_key,
extra,
privileges,
column_comment
FROM information_schema.columns AS i2
WHERE table_schema=db2 AND table_name=tbl2
) AS tmp
GROUP BY column_name
HAVING COUNT(*) = 1
ORDER BY column_name ;
|
DELIMITER ;
Compare two databases
One of EF Codd's rules for relational databases is the no-back-door rule: all info about tables should be accessible only by a query on tables. Since version 5, the MySQL implementation of
information_schema (I_S) helps meet Codd's requirement. I_S supplies metadata in tables, so it's the first place to look for how to compare the structures of two databases.
Elsewhere on this page there is a simple query template for comparing data in two structurally similar tables:
SELECT MIN(TableName) as TableName, id, col1, col2, col3, ...
FROM (
SELECT 'Table a' as TableName, a.id, a.col1, a.col2, a.col3, ...
FROM a
UNION ALL
SELECT 'Table b' as TableName, b.id, b.col1, b.col2, b.col3, ...
FROM b
) AS tmp
GROUP BY id, col1, col2, col3, ...
HAVING COUNT(*) = 1
ORDER BY ID;
To apply this logic to the comparison of two database structures:
- write temp tables collecting desired I_S metadata on each database
- map the compare-data query template to those two metadata tables
This logic is easiest to re-use when it is parameterised in a stored procedure, in a
system database:
USE sys;
DROP PROCEDURE IF EXISTS CompareDBs;
DELIMITER |
CREATE PROCEDURE CompareDBs( vdb1 VARCHAR(64), vdb2 VARCHAR(64) )
BEGIN
DROP TEMPORARY TABLE IF EXISTS desc1,desc2;
CREATE TEMPORARY TABLE desc1
SELECT
t1.table_schema,
t1.table_name,
t1.table_type,
t1.engine,
c1.column_name,
c1.ordinal_position,
c1.column_type,
c1.column_default,
c1.is_nullable,
c1.column_key
FROM information_schema.tables t1
JOIN information_schema.columns c1 USING (table_schema,table_name)
WHERE t1.table_schema=vdb1
ORDER BY t1.table_name,c1.column_name;
CREATE TEMPORARY TABLE desc2
SELECT
t1.table_schema,
t1.table_name,
t1.table_type,
t1.engine,
c1.column_name,
c1.ordinal_position,
c1.column_type,
c1.column_default,
c1.is_nullable,
c1.column_key
FROM information_schema.tables t1
JOIN information_schema.columns c1 USING (table_schema,table_name)
WHERE t1.table_schema=vdb2
ORDER BY t1.table_name,c1.column_name;
SELECT
TableName,column_name,MIN(SchemaName),table_type,engine,
ordinal_position,column_type,column_default,is_nullable,column_key
FROM (
SELECT
a.table_schema AS SchemaName,a.table_name AS TableName,a.table_type,a.engine,
a.column_name,a.ordinal_position,a.column_type,a.column_default,a.is_nullable,a.column_key
FROM desc1 a
UNION ALL
SELECT
b.table_schema AS SchemaName,b.table_name AS TableName,b.table_type,b.engine,
b.column_name,b.ordinal_position,b.column_type,b.column_default,b.is_nullable,b.column_key
FROM desc2 b
) AS tmp
GROUP BY TableName,table_type,engine,column_name,ordinal_position,column_type,column_default,is_nullable,column_key
HAVING COUNT(*) = 1
ORDER BY TableName,column_name,SchemaName;
DROP TEMPORARY TABLE desc1, desc2;
END |
DELIMITER ;
Call it as follows:
CALL compareDBs('db1','db2');
MEMORY tables would it speed it up, but unfortunately MySQL
MEMORY tables do not support
BLOB/TEXT columns.
Find child tables
Starting with MySQL 5, you can find all tables which have foreign key references to a given table with an
information_schema query, here encapsulated in a stored procedure which takes two arguments, a database name and a table name. When the table argument is blank or
NULL, the procedure returns all parent-child links where the parent table is in the specified database; otherwise it returns parent-child links for the specified parent table:
DROP PROCEDURE IF EXISTS ListChildren;
DELIMITER |
CREATE PROCEDURE ListChildren( pdb CHAR(64), ptbl CHAR(64) )
BEGIN
IF ptbl = '' OR ptbl IS NULL THEN
SELECT
c.table_schema as 'Parent Schema',
u.referenced_table_name as 'Parent Table',
u.referenced_column_name as 'Parent Column',
u.table_schema as 'Child Schema',
u.table_name as 'Child Table',
u.column_name as 'Child Column'
FROM information_schema.table_constraints AS c
JOIN information_schema.key_column_usage AS u USING(constraint_schema,constraint_name)
WHERE c.constraint_type = 'FOREIGN KEY'
AND u.referenced_table_schema = pdb
ORDER BY
c.table_schema, u.referenced_table_name, u.referenced_column_name,
u.table_schema, u.table_name, u.column_name;
ELSE
SELECT
c.table_schema as 'Parent Schema',
u.referenced_table_name as 'Parent Table',
u.referenced_column_name as 'Parent Column',
u.table_schema as 'Child Schema',
u.table_name as 'Child Table',
u.column_name as 'Child Column'
FROM information_schema.table_constraints AS c
JOIN information_schema.key_column_usage AS u USING(constraint_schema,constraint_name)
WHERE c.constraint_type = 'FOREIGN KEY'
AND u.referenced_table_schema = pdb
AND u.referenced_table_name = ptbl
ORDER BY
c.table_schema, u.referenced_table_name, u.referenced_column_name,
u.table_schema, u.table_name, u.column_name;
END IF;
END |
DELIMITER ;
Unfortunately it performs slowly, not because of how it is written, but because of how MySQL has implemented
information_schema. Indeed a
bug report page is devoted to the issue. It says MySQL AB will not be speeding up
information_schema query performance any time soon.
In
theUsual we worked around this problem with a PHP function that queries
information_schema if that is required, but by default parses the results of iterative
SHOW TABLES commands. The
SHOW TABLES method logic is simple, and will port readily to another application language. It runs 10-50 times faster than the equivalent
information_schema query.
The following version looks for child tables in one database; it slows down a bit when modified to search all server DBs, but even then it is much faster than its
information_schema equivalent. It assumes an available connection object
$conn:
function childtables( $db, $table, $via_infoschema=FALSE ) {
GLOBAL $conn;
$ret = array();
if( $via_infoschema ) {
$res = mysql_query( childtablesqry( $db, $table )) || die( mysql_error() );
if( !is_bool( $res ))
while( $row = mysql_fetch_row( $res ))
$ret[] = $row;
}
else {
$tables = array();
$res = mysql_query( "SHOW TABLES" );
while( $row = mysql_fetch_row( $res )) $tables[] = $row[0];
$res = mysql_query( "SELECT LOCATE('ANSI_QUOTES', @@sql_mode)" );
$ansi_quotes = $res ? mysql_result( $res, 0 ) : 0;
$q = $ansi_quotes ? '"' : "`";
$sref = ' REFERENCES ' . $q . $table . $q . ' (' . $q;
foreach( $tables as $referringtbl ) {
$res = mysql_query( "SHOW CREATE TABLE $referringtbl" );
$row = mysql_fetch_row( $res );
if(( $startref = stripos( $row[1], $sref )) > 0 ) {
$endref = strpos( $row[1], $q, $startref + strlen( $sref ));
$referencedcol = substr( $row[1], $startref+strlen($sref),
$endref-$startref-strlen($sref) );
$endkey = $startref;
while( substr( $row[1], $endkey, 1 ) <> $q ) $endkey--;
$startkey = --$endkey;
while( substr( $row[1], $startkey, 1 ) <> $q ) $startkey--;
$referencingcol = substr( $row[1], $startkey+1, $endkey - $startkey );
$ret[] = array( $db, $referringtbl, $referencingcol, $referencedcol );
}
}
}
return $ret;
}
function childtablesqry( $db, $table ) {
return "SELECT c.table_schema,u.table_name,u.column_name,u.referenced_column_name " .
"FROM information_schema.table_constraints AS c " .
"INNER JOIN information_schema.key_column_usage AS u " .
"USING( constraint_schema, constraint_name ) " .
"WHERE c.constraint_type = 'FOREIGN KEY' " .
"AND u.referenced_table_schema='$db' " .
"AND u.referenced_table_name = '$table' " .
"ORDER BY c.table_schema,u.table_name";
}
Find parent tables
List tables which are referenced by foreign key constraints in a given table.This is a simple query on two information_schema tables: table_constraints and key_column_usage. It is easy to parameterise, so we show it in stored procedures. The first sproc lists all foreign key references in a database. The second lists all foreign key references for a table.
CREATE PROCEDURE ListParentsForDb( pdb CHAR(64) )
BEGIN
SELECT
u.table_schema AS 'Schema',
u.table_name AS 'Table',
u.column_name AS 'Key',
u.referenced_table_schema AS 'Parent Schema',
u.referenced_table_name AS 'Parent table',
u.referenced_column_name AS 'Parent key'
FROM information_schema.table_constraints AS c
INNER JOIN information_schema.key_column_usage AS u
USING( constraint_schema, constraint_name )
WHERE c.constraint_type = 'FOREIGN KEY'
AND c.table_schema = pdb
ORDER BY u.table_schema,u.table_name,u.column_name;
END;
CREATE PROCEDURE ListParentsForTable( pdb CHAR(64), ptable CHAR(64) )
BEGIN
SELECT
u.table_schema AS 'Schema',
u.table_name AS 'Table',
u.column_name AS 'Key',
u.referenced_table_schema AS 'Parent Schema',
u.referenced_table_name AS 'Parent table',
u.referenced_column_name AS 'Parent key'
FROM information_schema.table_constraints AS c
INNER JOIN information_schema.key_column_usage AS u
USING( constraint_schema, constraint_name )
WHERE c.constraint_type = 'FOREIGN KEY'
AND c.table_schema = pdb
AND u.referenced_table_name = ptable
ORDER BY u.table_schema,u.table_name,u.column_name;
END;
Find primary key of a table
To retrieve primary keys of db.tbl...
SELECT k.column_name
FROM information_schema.table_constraints t
JOIN information_schema.key_column_usage k
USING (constraint_name,table_schema,table_name)
WHERE t.constraint_type='PRIMARY KEY'
AND t.table_schema='db'
AND t.table_name='tbl'
For pre-5 versions of MySQL:
SHOW INDEX FROM tbl
WHERE key_name='primary';
Find the size of all databases on the server
This is based on a query Mark Leith posted to the MySQL General Discussion list.
DROP VIEW IF EXISTS dbsize;
CREATE VIEW dbsize AS
SELECT
s.schema_name AS 'Schema',
SUM(t.data_length) AS Data,
SUM( t.index_length ) AS Indexes,
SUM(t.data_length) + SUM(t.index_length) AS 'Mb Used',
IF(SUM(t.data_free)=0,'',SUM(t.data_free)) As 'Mb Free',
IF( SUM(t.data_free)=0,
'',
100 * (SUM(t.data_length) + SUM(t.index_length)) / ((SUM(t.data_length)+SUM(t.index_length) + SUM(IFNULL(t.data_free,0))) )
) AS 'Pct Used',
COUNT(table_name) AS Tables
FROM information_schema.schemata s
LEFT JOIN information_schema.tables t ON s.schema_name = t.table_schema
GROUP BY s.schema_name
WITH ROLLUP
List databases, tables, columns
SELECT
t.table_schema AS 'Database'
,t.table_name AS 'Table'
,t.table_type AS 'Table Type'
,c.column_name AS 'Column'
,c.data_type AS 'Data Type'
FROM information_schema.tables t
JOIN information_schema.columns c ON t.table_schema = c.table_schema AND t.table_name = c.table_name
WHERE t.table_schema NOT IN( 'mysql','information_schema')
ORDER BY t.table_schema,t.table_type,t.table_name,c.ordinal_position;
List differences between two databases
Given two databases named @db1 and @db2:
SELECT
MIN(table_name) as TableName,
table_catalog,table_schema,table_name,column_name,
ordinal_position,column_default,is_nullable,
data_type,character_maximum_length,character_octet_length,
numeric_precision,numeric_scale,character_set_name,
collation_name,column_type,column_key,
extra,privileges,column_comment
FROM (
SELECT 'Table a' as TableName,
table_catalog,table_schema,table_name,column_name,
ordinal_position,column_default,is_nullable,
data_type,character_maximum_length,character_octet_length,
numeric_precision,numeric_scale,character_set_name,
collation_name,column_type,column_key,
extra,privileges,column_comment
FROM information_schema.columns c1
WHERE table_schema=@db1
UNION ALL
SELECT 'Table a' as TableName,
table_catalog,table_schema,table_name,column_name,
ordinal_position,column_default,is_nullable,
data_type,character_maximum_length,character_octet_length,
numeric_precision,numeric_scale,character_set_name,
collation_name,column_type,column_key,
extra,privileges,column_comment
FROM information_schema.columns c2
WHERE table_schema=@db2
) AS tmp
GROUP BY tablename,
table_catalog,table_schema,column_name,
ordinal_position,column_default,is_nullable,
data_type,character_maximum_length,character_octet_length,
numeric_precision,numeric_scale,character_set_name,
collation_name,column_type,column_key,
extra,privileges,column_comment
HAVING COUNT(*) = 1
ORDER BY tablename,column_name;
List users of a database
DROP PROCEDURE IF EXISTS ListDbUsers;
DELIMITER |
CREATE PROCEDURE ListDbUsers( dbname CHAR(64) )
SELECT host,user
FROM mysql.user
WHERE Select_priv = 'Y'
OR Insert_priv = 'Y'
OR Update_priv = 'Y'
OR Delete_priv = 'Y'
OR Create_priv = 'Y'
OR Drop_priv = 'Y'
OR Reload_priv = 'Y'
OR Shutdown_priv = 'Y'
OR Process_priv = 'Y'
OR File_priv = 'Y'
OR Grant_priv = 'Y'
OR References_priv = 'Y'
OR Index_priv = 'Y'
OR Alter_priv = 'Y'
OR Show_db_priv = 'Y'
OR Super_priv = 'Y'
OR Create_tmp_table_priv = 'Y'
OR Lock_tables_priv = 'Y'
OR Execute_priv = 'Y'
OR Repl_slave_priv = 'Y'
OR Repl_client_priv = 'Y'
OR Create_view_priv = 'Y'
OR Show_view_priv = 'Y'
OR Create_routine_priv = 'Y'
OR Alter_routine_priv = 'Y'
OR Create_user_priv = 'Y'
OR Event_priv = 'Y'
OR Trigger_priv = 'Y'
UNION
SELECT host,user
FROM mysql.db
WHERE db=dbname
AND (
Select_priv = 'Y'
OR Insert_priv = 'Y'
OR Update_priv = 'Y'
OR Delete_priv = 'Y'
OR Create_priv = 'Y'
OR Drop_priv = 'Y'
OR Grant_priv = 'Y'
OR References_priv = 'Y'
OR Index_priv = 'Y'
OR Alter_priv = 'Y'
OR Create_tmp_table_priv = 'Y'
OR Lock_tables_priv = 'Y'
OR Create_view_priv = 'Y'
OR Show_view_priv = 'Y'
OR Create_routine_priv = 'Y'
OR Alter_routine_priv = 'Y'
OR Execute_priv = 'Y'
OR Event_priv = 'Y'
OR Trigger_priv = 'Y'
)
UNION
SELECT host,user
FROM mysql.tables_priv
WHERE db=dbname
UNION
SELECT host,user
FROM mysql.columns_priv
WHERE db=dbname;
|
DELIMITER ;
CALL ListDbUsers( 'test' );
Rename Database
It's sometimes necessary to rename a database. MySQL 5.0 has no command for it. Simply bringing down the server to rename a database directory is
not safe. MySQL 5.1.7 introduced a
RENAME DATABASE command, but the command left several unchanged database objects behind, and was found to lose data, so it was dropped in 5.1.23.
It seems a natural for a stored procedure using dynamic (prepared) statements.
PREPARE supports
CREATE | RENAME TABLE. As precautions:
- Before calling the sproc, the new database must have been created.
- The procedure refuses to rename the mysql database.
- The old database is left behind, minus what was moved.
DROP PROCEDURE IF EXISTS RenameDatabase;
DELIMITER |
CREATE PROCEDURE RenameDatabase(
IN oldname CHAR (64), IN newname CHAR(64)
)
BEGIN
DECLARE version CHAR(32);
DECLARE sname CHAR(64) DEFAULT NULL;
DECLARE rows INT DEFAULT 1;
DECLARE changed INT DEFAULT 0;
IF STRCMP( oldname, 'mysql' ) <> 0 THEN
REPEAT
SELECT table_name INTO sname
FROM information_schema.tables AS t
WHERE t.table_type='BASE TABLE'
AND t.table_schema = oldname
LIMIT 1;
SET rows = FOUND_ROWS();
IF rows = 1 THEN
SET @scmd = CONCAT( 'RENAME TABLE ', oldname, '.', sname,
' TO ', newname, '.', sname );
PREPARE cmd FROM @scmd;
EXECUTE cmd;
DEALLOCATE PREPARE cmd;
SET changed = 1;
END IF;
UNTIL rows = 0 END REPEAT;
IF changed > 0 THEN
SET @scmd = CONCAT( "UPDATE mysql.db SET Db = '",
newname,
"' WHERE Db = '", oldname, "'" );
PREPARE cmd FROM @scmd;
EXECUTE cmd;
DROP PREPARE cmd;
SET @scmd = CONCAT( "UPDATE mysql.proc SET Db = '",
newname,
"' WHERE Db = '", oldname, "'" );
PREPARE cmd FROM @scmd;
EXECUTE cmd;
DROP PREPARE cmd;
SELECT version() INTO version;
IF version >= '5.1.7' THEN
SET @scmd = CONCAT( "UPDATE mysql.event SET db = '",
newname,
"' WHERE db = '", oldname, "'" );
PREPARE cmd FROM @scmd;
EXECUTE cmd;
DROP PREPARE cmd;
END IF;
SET @scmd = CONCAT( "UPDATE mysql.columns_priv SET Db = '",
newname,
"' WHERE Db = '", oldname, "'" );
PREPARE cmd FROM @scmd;
EXECUTE cmd;
DROP PREPARE cmd;
FLUSH PRIVILEGES;
END IF;
END IF;
END;
|
DELIMITER ;
Show Create Trigger
MySQL added a Show Create Trigger command in 5.1.21. If you use an earlier MySQL version, here is a stored procedure which behaves like Show Create Trigger:
DROP PROCEDURE IF EXISTS ShowCreateTrigger;
DELIMITER go
CREATE PROCEDURE ShowCreateTrigger( IN db CHAR(64), IN tbl CHAR(64) )
BEGIN
SELECT
CONCAT(
'CREATE TRIGGER ',trigger_name, CHAR(10),
action_timing,' ', event_manipulation, CHAR(10),
'ON ',event_object_schema,'.',event_object_table, CHAR(10),
'FOR EACH ROW', CHAR(10),
action_statement, CHAR(10)
) AS 'Triggers'
FROM information_schema.triggers
WHERE event_object_schema = db
AND event_object_table = tbl;
END;
go
DELIMITER ;
And here is a stored procedure which lists all triggers in a database:
DROP PROCEDURE IF EXISTS ListTriggers;
DELIMITER go
CREATE PROCEDURE ListTriggers( IN db CHAR(64) )
BEGIN
SELECT
trigger_name AS 'Trigger',
event_object_table AS 'Table'
FROM information_schema.triggers
WHERE event_object_schema = db;
END;
go
DELIMITER ;
If you have a collection of generic stored procs like these, it's most convenient to keep them in one place for easy accessibility. We keep ours in a sys database.
Last updated 09 Oct 2009
Show Table Status equivalent from information_schema
Fill in schema and table names in ...
SELECT
table_name,
engine,
version,
row_format,
table_rows,
avg_row_length,
data_length,
max_data_length,
index_length,
data_free,
auto_increment,
create_time,
update_time,
check_time,
table_collation,
checksum,
create_options,
table_comment
FROM information_schema.tables
where table_schema='???' AND table_name='???';
Last updated 09 Oct 2009
Show Tables
The MySQL SHOW TABLES command is fine, but sometimes we want a little more information.
This simple stored procedure lists the table name, engine type, version, collation and rowcount for every table in a database. (Individual databases come and go, so we keep all such database-wide stored routines in a system database.)
DROP PROCEDURE IF EXISTS showtables;
CREATE PROCEDURE showtables()
SELECT
table_name AS 'Table',
IFNULL(engine, 'VIEW') AS Engine,
version AS Version,
table_collation AS Collation,
table_rows AS Rows
FROM information_schema.tables
WHERE table_schema=DATABASE();
If you often want to list tables in a DB not currently USEd, here is a parameter-driven version:
DROP PROCEDURE IF EXISTS showtablesDB;
CREATE PROCEDURE showtablesDB( IN dbname CHAR(64) )
SELECT
table_name AS 'Table',
IFNULL(engine, 'VIEW') AS Engine,
version AS Version,
table_collation AS Collation,
table_rows AS Rows
FROM information_schema.tables
WHERE table_schema=dbname;
For a sys database of ours, this shows ...
+-------------+--------+---------+-------------------+------+
| Table | Engine | Version | Collation | Rows |
+-------------+--------+---------+-------------------+------+
| dbsize | VIEW | NULL | NULL | NULL |
| dcal | MEMORY | 10 | latin1_swedish_ci | 0 |
| digits | MyISAM | 10 | latin1_swedish_ci | 10 |
| errors | MEMORY | 10 | latin1_swedish_ci | 0 |
| ints | MyISAM | 10 | latin1_swedish_ci | 10 |
| results | MEMORY | 10 | latin1_swedish_ci | 0 |
| theusualfks | MyISAM | 10 | latin1_swedish_ci | 5 |
| viewparams | MyISAM | 10 | latin1_swedish_ci | 2 |
+-------------+--------+---------+-------------------+------+
Last updated 22 Nov 2009
Age in years
Given a birthdate in @dob, here is a simple formula for age in years:
DATE_FORMAT(FROM_DAYS(TO_DAYS(now()) - TO_DAYS(@dob)), '%Y') + 0;
and here is one for age in years to two decimal places (ignoring day of month):
ROUND((((YEAR(now()) - YEAR(@dob)))*12 + (((MONTH(now()) - MONTH(@dob)))))/12, 2)
Appointments available
Given a clinic of physicians, patients and appointments, how to find an available appointment time for a given physician?
This is a variant of the [Not] Exists query pattern. Though we can write it with subqueries, performance will be crisper with a join. But finding data that is not there requires a join to data which is there. So in addition to tables for appointments, doctors and patients, we need a table of all possible appointment datetimes. Here's a schema illustrating the idea ...
CREATE TABLE a_dt ( -- POSSIBLE APPOINTMENT DATES AND TIMES
d DATE,
t TIME
);
CREATE TABLE a_drs ( -- DOCTORS
did INT -- doctor id
);
CREATE TABLE a_pts ( -- PATIENTS
pid INT
);
CREATE TABLE a_appts ( -- APPOINTMENTS
aid INT, -- appt id
did INT, -- doctor id
pid INT, -- patient id
d DATE,
t TIME
);
Now we can apply the [Not] Exists query pattern. To find free appointment datetimes for a given doctor in a given datetime range, we left join possible appointments to existing appointments on date and time and doctor, add Where conditions for desired appointment datetimes, and finally add a Where condition that the appointment slot be null, i.e. free...
SELECT d.did, a.d, a.t
FROM a_dt AS a
LEFT JOIN a_appts AS ap USING (d,t)
LEFT JOIN a_drs AS d
ON a.d = ap.d
AND a.t = ap.t
AND ap.did = d.did
AND ap.did = 1
WHERE a.d BETWEEN desired_start_date AND desired_end_date
AND a.t BETWEEN desired_start_time AND desired_end_time
AND ap.aid IS NULL;
Count business days between two dates
The simplest support for counting business days between any two dates is a calendar table calendar(d date,isholiday bool) populated for all days in all possibly relevant years. Then the following query gives the inclusive number of business days between dates dStart and dStop:
SELECT COUNT(*)
FROM calendar
WHERE d BETWEEN dStart AND dStop
AND DAYOFWEEK(d) NOT IN(1,7)
AND isholiday=0;
If that solution is not available, you have to do with a weekday count, which this function (corrected 6 Jul 2009) computes:
SET GLOBAL log_bin_trust_function_creators=1;
DROP FUNCTION IF EXISTS BizDaysInclusive;
DELIMITER |
CREATE FUNCTION BizDaysInclusive( d1 DATE, d2 DATE )
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE dow1, dow2, days INT;
SET dow1 = DAYOFWEEK(d1);
SET dow2 = DAYOFWEEK(d2);
SET days = FLOOR( DATEDIFF(d2,d1)/7 ) * 5 +
CASE
WHEN dow1=1 AND dow2=7 THEN 5
WHEN dow1 IN(7,1) AND dow2 IN (7,1) THEN 0
WHEN dow1=dow2 THEN 1
WHEN dow1 IN(7,1) AND dow2 NOT IN (7,1) THEN dow2-1
WHEN dow1 NOT IN(7,1) AND dow2 IN(7,1) THEN 7-dow1
WHEN dow1<=dow2 THEN dow2-dow1+1
WHEN dow1>dow2 THEN 5-(dow1-dow2-1)
ELSE 0
END;
RETURN days;
END;
|
DELIMITER ;
SELECT BizDaysInclusive('2009-8-1','2009-9-15');
+------------------------------------------+
| BizDaysInclusive('2009-8-1','2009-9-15') |
+------------------------------------------+
| 32 |
+------------------------------------------+
So does the following algorithm, posted by Elowie Cruz on a MySQL forum; for ease of use we encapsulate it in a stored procedure:
DROP PROCEDURE IF EXISTS daycount;
CREATE PROCEDURE DayCount( d1 DATE, d2 DATE )
SELECT dd.iDiff, dd.iDiff - dd.iWeekEndDays AS iWorkDays, dd.iWeekEndDays
FROM (
SELECT
dd.iDiff,
((dd.iWeeks * 2) +
IF(dd.iSatDiff >= 0 AND dd.iSatDiff < dd.iDays, 1, 0) +
IF (dd.iSunDiff >= 0 AND dd.iSunDiff < dd.iDays, 1, 0)) AS iWeekEndDays
FROM (
SELECT
dd.iDiff,
FLOOR(dd.iDiff / 7) AS iWeeks,
dd.iDiff % 7 iDays,
5 - dd.iStartDay AS iSatDiff,
6 - dd.iStartDay AS iSunDiff
FROM (
SELECT
1 + DATEDIFF(d2, d1) AS iDiff,
WEEKDAY(d1) AS iStartDay
) AS dd
) AS dd
) AS dd ;
CALL DayCount( '2009-8-1','2009-9-15');
+-------+-----------+--------------+
| iDiff | iWorkDays | iWeekEndDays |
+-------+-----------+--------------+
| 46 | 32 | 14 |
+-------+-----------+--------------+
Last updated 10 Jul 2009
Count Tuesdays between two dates
Date arithmetic is deceptively hard. One way to appreciate the difficulties is to read
Chapter 21 in our book. Another is to try to calculate the number of Tuesdays (or another weekday) between two dates. It's not a back-of-the-napkin problem.
An earlier formula we had for this problem sometimes gave incorrect results. As a debugging aid, we wrote a brute force calculator for the problem:
SET GLOBAL log_bin_trust_function_creators=1;
DROP FUNCTION IF EXISTS DayCount;
DELIMITER |
CREATE FUNCTION DayCount( d1 DATE, d2 DATE, daynum SMALLINT ) RETURNS INT
BEGIN
DECLARE days INT DEFAULT 0;
IF D1 IS NOT NULL AND D2 IS NOT NULL THEN
WHILE D1 <= d2 DO
BEGIN
IF DAYOFWEEK(d1) = daynum THEN
SET days=days+1;
END IF;
SET d1 = ADDDATE(d1, INTERVAL 1 DAY);
END;
END WHILE;
END IF;
RETURN days;
END;
|
DELIMITER ;
select
daycount('2008-3-16','2008-3-28',7)
AS 'Sats from 2008-3-16 through 2008-3-28';
You would not want to use that function on long date spans in a big table, but it will do for testing.
Now, how to count the number of
Tuesdays, say, between two dates? The basic logic is:
1. Count
weeks between the two dates.
2. If beginning and ending weekdays are the same, then if they're Tuesday, the answer is
weeks+1, otherwise it's just
weeks.
3. Otherwise, if the beginning weekday <= the ending weekday, then if Tuesday is between them, the answer is
weeks+1, otherwise it's just
weeks.
4. Otherwise the ending weekday is less than the starting weekday; if Tuesday >= the starting weekday or <= the ending weekdayy, the answer is
weeks+1, otherwise it's just
weeks.
For a convenient datasource, we'll use the two date columns
orderdate and
shippeddate in the
orders table of the NorthWind database, and we'll use our brute force function DayCount() to check results:
SET @day = 3;
SELECT
DATE_FORMAT(orderdate,'%y%m%d') AS OrdDt,
DATE_FORMAT(shippeddate,'%y%m%d') AS ShipDt,
LEFT(DAYNAME(orderdate),3) AS D1,
LEFT(DAYNAME(shippeddate),3) AS D2,
@dow1 := DAYOFWEEK(orderdate) AS 'dw1',
@dow2 := DAYOFWEEK(shippeddate) AS 'dw2',
@days := DATEDIFF(shippeddate,orderdate) AS Days,
@wks := FLOOR( @days / 7 ) AS Wks,
FLOOR( IF( @dow1 = @dow2, IF( @day = @dow1, @wks+1, @wks),
IF( @dow1 < @dow2, IF( @day BETWEEN @dow1 AND @dow2, @wks+1, @wks),
IF( @day >= @dow1 OR @day <= @dow2, @wks+1, @wks )
)
)
) AS Res,
DayCount(DATE(orderdate),DATE(shippeddate),@day) AS Chk
FROM orders
HAVING !ISNULL(res-chk) AND res-chk <> 0;
Empty set (0.00 sec)
No errors. We get the same result for @day = 1, 2, 4, 5, 6 and 7.
But the formula is buried in the specifics of one table, so abstract it to a reusable function:
DROP FUNCTION IF EXISTS NamedDaysBetween;
DELIMITER |
CREATE FUNCTION NamedDaysBetween( d1 DATE, d2 DATE, daynum SMALLINT )
RETURNS INT
BEGIN
DECLARE dow1, dow2, wks, days INT;
IF !ISNULL(d1) AND !ISNULL(d2) THEN
SET dow1 = DAYOFWEEK( d1 );
SET dow2 = DAYOFWEEK( d2 );
SET days = DATEDIFF( d2, d1 );
SET wks = FLOOR( days / 7 );
SET days = IF( dow1 = dow2, IF( daynum = dow1, wks+1, wks),
IF( dow1 < dow2, IF( daynum BETWEEN dow1 AND dow2, wks+1, wks),
IF( daynum >= dow1 OR daynum <= dow2, wks+1, wks )
)
);
END IF;
RETURN days;
END;
|
DELIMITER ;
Again check it against lots of date value pairs:
SELECT
nameddaysbetween(orderdate,shippeddate,3) - daynamecount(orderdate,shippeddate,3)
AS diff
FROM orders
HAVING !ISNULL(diff) AND diff <> 0;
Empty set (0.00 sec)
Date of first Friday of next month
Assuming a calendar table calendar(date DATE) with one row per date through the relevant period...
SET @d = NOW();
SELECT MIN(date) AS 1stFridayOfMonth
FROM calendar
WHERE YEAR(date) = IF( MONTH(@d) = 12, 1+YEAR(@d), YEAR(@d) )
AND MONTH(date) = IF( MONTH(@d) = 12, 1, MONTH(@d) + 1 )
AND WEEKDAY(date)=4;
Date of Monday in a given week of the year
The week number of a given date @d, when weeks are defined as starting on Mondays and when we agree to number weeks of the year from 1 through 53, is given by WEEK(@d, 2). Here is a way to get the date of Monday in that week:
set @d='2008-1-31';
select makedate( left(yearweek(@d),4),week( @d, 2 ) * 7 ) as 1stdayOfWeek;
+--------------+
| 1stdayOfWeek |
+--------------+
| 2008-01-28 |
+--------------+
set @d='2008-7-15';
select makedate( left(yearweek(@d),4),week( @d, 2 ) * 7 ) as 1stdayOfWeek;
+--------------+
| 1stdayOfWeek |
+--------------+
| 2008-07-14 |
+--------------+
Datetime difference
Find the difference between two datetime values in seconds, minutes, hours or days. If dt1 and dt2 are datetime values of the form 'yyyy-mm-dd hh:mm:ss', the number of seconds between dt1 and dt2 is
UNIX_TIMESTAMP( dt2 ) - UNIX_TIMESTAMP( dt1 )
To get the number of minutes divide by 60, for the number of hours divide by 3600, and for the number of days, divide by 3600 * 24.
Duration in years, months, days and time
DROP FUNCTION IF EXISTS PeriodLen;
DELIMITER |
CREATE FUNCTION PeriodLen( dt1 datetime, dt2 datetime ) RETURNS CHAR(128)
BEGIN
DECLARE yy,mm,d0,dd,hh,mi,ss,t1 BIGINT;
DECLARE t0 TIMESTAMP;
SET yy = TIMESTAMPDIFF(YEAR,dt1,dt2);
SET mm = TIMESTAMPDIFF(MONTH,dt1,dt2) MOD 12;
SET d0 = TIMESTAMPDIFF(DAY,dt1,dt2);
SET dd = d0 MOD (yy*mm);
SET t0 = TIMESTAMPADD(DAY,d0,dt1);
SET t1 = TIME_TO_SEC(TIMEDIFF(dt2,t0));
SET hh = FLOOR(t1/3600);
SET mi = FLOOR(t1/60) - 60*hh;
SET ss = t1 - 3600*hh - 60*mi;
RETURN CONCAT( yy, ' years ', mm, ' months ', dd, ' days ', hh, ' hours ', mi, ' mins ', ss, ' secs' );
END |
DELIMITER ;
SELECT PeriodLen( '2000-6-1 20:34:35',now() );
+----------------------------------------------------+
| PeriodLen( '2000-6-1 20:34:35',now() ) |
+----------------------------------------------------+
| 8 years 11 months 19 days 17 hours 49 mins 22 secs |
+----------------------------------------------------+
Last updated 22 May 2009
Find available reservation periods
Given a bookings table where each row specifies one reservation period for one property, find the unbooked periods for a given property:
CREATE TABLE bookings( ID int, propertyID int, startDate date, endDate date );
INSERT INTO bookings VALUES
(1,1,'2007-1-1','2007-1.15'),
(2,1,'2007-1-20','2007-1.31'),
(3,1,'2007-2-10','2007-2-17');
SELECT * FROM bookings;
+------+------------+------------+------------+
| ID | propertyID | startDate | endDate |
+------+------------+------------+------------+
| 1 | 1 | 2007-01-01 | 2007-01-15 |
| 2 | 1 | 2007-01-20 | 2007-01-31 |
| 3 | 1 | 2007-02-10 | 2007-02-17 |
+------+------------+------------+------------+
Reservation systems usually adopt the closed-open convention of representing when reservations begin and end. For example, if you book a hotel room for 22 May through 24 May, the hotel expects you to stay overnight on 22 May and 23 May, but not on 24 May. Apart from that difference, this is the same pattern as Finding missing numbers in a sequence.
SELECT
a.enddate AS 'Available From',
Min(b.startdate) AS 'To'
FROM bookings AS a
JOIN bookings AS b ON a.propertyID=b.propertyID AND a.enddate < b.startdate
WHERE a.propertyID=1
GROUP BY a.enddate
HAVING a.enddate < MIN(b.startdate);
+----------------+------------+
| Available From | To |
+----------------+------------+
| 2007-01-15 | 2007-01-20 |
| 2007-01-31 | 2007-02-10 |
+----------------+------------+
This query cannot see reservation dates earlier than the first existing reservation date, or later than the last. Usually, you would want a calendar table to provide those limits, but you can fake them with a union. If the allowable reservation period is 1 Dec 2006 through 1 Jul 2007, union the left side of the join with a made-up row for 1 Dec 2006, and union the right side of the join with a made-up row for 1 Jul 2007:
SELECT
a.enddate AS 'Available From',
Min(b.startdate) AS 'To'
FROM (
SELECT 0,1 as propertyID,'2006-12-01' as startdate,'2006-12-01' as enddate
UNION
SELECT * FROM bookings
) AS a
JOIN (
SELECT * FROM bookings
UNION
SELECT 0,1,'2007-07-01' as startdate,'2007-07-02' as enddate
) AS b ON a.propertyID=b.propertyID AND a.enddate < b.startdate
WHERE a.propertyID=1
GROUP BY a.enddate
HAVING a.enddate < MIN(b.startdate);
+----------------+------------+
| Available From | To |
+----------------+------------+
| 2006-12-01 | 2007-01-01 |
| 2007-01-15 | 2007-01-20 |
| 2007-01-31 | 2007-02-10 |
| 2007-02-17 | 2007-07-01 |
+----------------+------------+
Find overlapping periods
You have a table of visits, and you would like to display the time periods during which there are visit time overlaps.
drop table if exists visits;
create table visits(id int primary key,start datetime,end datetime);
insert into visits values
(1, '2008-09-01 15:01', '2008-09-01 15:04'),
(2, '2008-09-01 15:02', '2008-09-01 15:09'),
(3, '2008-09-01 15:12', '2008-09-01 15:15'),
(4, '2008-09-01 16:11', '2008-09-01 16:23'),
(5, '2008-09-01 16:19', '2008-09-01 16:25'),
(6, '2008-09-01 17:52', '2008-09-01 17:59'),
(7, '2008-09-01 18:18', '2008-09-01 18:22'),
(8, '2008-09-01 16:20', '2008-09-01 16:22'),
(9, '2008-09-01 18:17', '2008-09-01 18:23');
There are five period overlaps in this data:
1 |-----|
2 |-----|
3 |--|
4 |-----|
5 |-----|
6 |---|
7 |---|
8 |---|
9 |-----|
One solution is to use a View to identify starting and stopping events, then define an Overlaps View:
CREATE OR REPLACE VIEW events AS
SELECT start AS time, 1 AS value, id FROM visits
UNION
SELECT end AS time, -1 AS value, id FROM visits;
CREATE OR REPLACE VIEW overlaps AS
SELECT time t, (SELECT SUM(value) FROM events WHERE time <=t ) as visitcount
FROM events;
SELECT t, visitcount
FROM overlaps
WHERE visitcount>1
ORDER BY t;
+---------------------+------------+
| t | visitcount |
+---------------------+------------+
| 2008-09-01 15:02:00 | 2 |
| 2008-09-01 16:19:00 | 2 |
| 2008-09-01 16:20:00 | 3 |
| 2008-09-01 16:22:00 | 2 |
| 2008-09-01 18:18:00 | 2 |
+---------------------+------------+
But that doesn't show us when overlap periods end. There is a fuller and more straightforward solution: join visits to itself on the criteria that
(i) the first of each joined pair of visits started no earlier than the second,
(ii) the first visit started before the second ended, and
(iii) the second visit started before the first ended:
Thanks to Brian Moretta for pointing out the need for LEAST() to catch all endpoints:
SELECT v1.id, v1.start, v2.id, LEAST(v1.end,v2.end) AS end
FROM visits v1
JOIN visits v2 ON v1.id <> v2.id and v1.start >= v2.start and v1.start < v2.end
ORDER BY v1.start;
+----+---------------------+----+---------------------+
| id | start | id | end |
+----+---------------------+----+---------------------+
| 2 | 2008-09-01 15:02:00 | 1 | 2008-09-01 15:04:00 |
| 5 | 2008-09-01 16:19:00 | 4 | 2008-09-01 16:23:00 |
| 8 | 2008-09-01 16:20:00 | 4 | 2008-09-01 16:22:00 |
| 8 | 2008-09-01 16:20:00 | 5 | 2008-09-01 16:22:00 |
| 7 | 2008-09-01 18:18:00 | 9 | 2008-09-01 18:22:00 |
+----+---------------------+----+---------------------+
Last updated 05 Sep 2009
Find sequenced duplicates
A table that tracks time periods may require period uniqueness. That means it has no sequenced duplicates.
If a table has columns processID, start_date and end_date, those three columns are period unique if there exists no pair of rows with the same processID and overlapping start_date and end_date values. If there is such a pair of rows, the table exhibits sequenced duplication.
Another way of saying it: if an instant is the smallest datetime unit of start_date and end_date columns, then if there are no sequenced duplicates, there is exactly one processID value at any instant.
Here is a query to find sequenced duplicates for those columns:
SELECT t.processid
FROM tbl t
WHERE EXISTS (
SELECT * FROM tbl AS t3
WHERE t3.processid IS NULL
)
OR EXISTS (
SELECT * FROM tbl AS t1
WHERE 1 < (
SELECT COUNT(processid)
FROM tbl AS t2
WHERE t1.processid = t2.processid
AND t1.start_date < t2.end_date
AND t2.start_date < t1.end_date
)
);
Is a given booking period available?
You rent vacation properties, tracking bookings with a table like this:
CREATE TABLE bookings( ID int, propertyID int, startDate date, endDate date );
INSERT INTO bookings VALUES (1,1,'2007-1-1','2007-1.15'),(2,1,'2007-1-20','2007-1.31');
SELECT * FROM bookings;
+------+------------+------------+------------+
| ID | propertyID | startDate | endDate |
+------+------------+------------+------------+
| 1 | 1 | 2007-01-01 | 2007-01-15 |
| 2 | 1 | 2007-01-20 | 2007-01-31 |
+------+------------+------------+------------+
You need a query indicating whether a given property is available for a given period of time.
Hotels & property renters usually adopt what is called the 'closed-open' convention for bookings, eg a booking from 22 May through 24 May means you sleep there the nights of 22 and 23 May. To show that property P is available for the desired closed-open period dStart to dEnd, you need to prove there is no booked period for P that overlaps dStart through dEnd. Until you're used to thinking about periods, it's easier to analyse graphically. There are four ways a booked reservation can overlap the desired date range ...
dStart dEnd
|----------------|
startDate endDate
|--------------------------|
|------|
|----|
|------|
but there are just two ways a booked reservation can not overlap:
dStart dEnd
|----------------|
|-----| |-----|
|-----| |-----|
So the period dStart through dEnd is available if there is no row where ...
!(endDate <= dStart OR startDate >= dEnd)
or equivalently ...
endDate > dStart AND startDate < dEnd
Here is a simple stored procedure for testing the query:
DROP PROCEDURE IF EXISTS isavailable;
DELIMITER |
CREATE PROCEDURE isavailable( iProperty int, dStart date, dEnd date )
SELECT IF( COUNT(1),'No','Yes' ) AS Available
FROM bookings
WHERE propertyID = iProperty
AND startDate < dEnd
AND endDate > dStart;
|
DELIMITER ;
CALL isavailable(1,'2006-12-27','2007-1-20');
+-----------+
| Available |
+-----------+
| No |
+-----------+
CALL isavailable(1,'2007-1-10' ,'2007-1-16');
+-----------+
| Available |
+-----------+
| No |
+-----------+
CALL isavailable(1,'2007-1-16' ,'2007-1-17');
+-----------+
| Available |
+-----------+
| Yes |
+-----------+
CALL isavailable(1,'2007-1-22' ,'2007-1-23');
+-----------+
| Available |
+-----------+
| No |
+-----------+
CALL isavailable(1,'2007-1-22' ,'2007-2-2');
+-----------+
| Available |
+-----------+
| No |
+-----------+
CALL isavailable(1,'2007-2-1' ,'2007-2-2');
+-----------+
| Available |
+-----------+
| Yes |
+-----------+
CALL isavailable(1,'2006-12-1' ,'2007-2-1');
+-----------+
| Available |
+-----------+
| No |
+-----------+
1 row in set (0.00 sec)
Julian date
Unix_Timestamp( datetimevalue ) / (60*60*24) ) + 2440587.5
Last business day before a reference date
Given a date value in datetimecol ...
SELECT
@refday := datetimecol,
@dow := DAYOFWEEK(@refday) AS DOW,
@subtract := IF( @dow = 1, 2, IF( @dow = 2, 3, 1 )) AS MINUS,
@refday - INTERVAL @subtract DAY AS LastBizDay
FROM ... etc
Make a calendar table
You need a calendar table for joins to datetime data in other tables:
create table calendar ( dt datetime primary key );
An elegant method of generating any desired number of sequential values, posted by Giuseppe Maxia on his
blog, is ...
- Create three dummy rows in a View.
- Cross join them to make 10 dummy rows.
- Cross join those to make 100, 1,000 or however many you need.
So to give the calendar table a million rows at one-hour intervals starting on 1 Jan 1970:
create view v3 as select 1 n union all select 1 union all select 1;
create view v as select 1 n from v3 a, v3 b union all select 1;
set @n = 0;
insert into calendar select '1970-1-1 00:00:00'+interval @n:=@n+1 hour
from v a, v b, v c, v d, v e, v;
If you prefer to do it one query without user variables, or if your MySQL version does not support Views, make a general-purpose utility table of integers 0 through 9. We keep such utility objects in a
sys database and make it available to all users:
create table sys.ints(i tinyint);
insert into sys.ints values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
From
sys.ints you can generate any desired sequence of dates or datetimes. For example to report monthly sales totals for 2009 from the table
sales(date d, amount decimal(10,2)), write:
select a.yearmonth, sum(ifnull(b.amount,0)) as total
from (
select concat( '2009-', lpad((u.i + t.i * 10), 2, 0)) as yearmonth
from sys.ints u, sys.ints t
where (u.i + t.i * 10) between 1 and 12
) a
left join sales b on a.yearmonth=date_format(b.d,'%Y-%m')
group by a.yearmonth;
Generate a thousand daily dates starting today:
SELECT CURDATE() + INTERVAL t.i*100 + u.i*10 + v.i DAY AS Date
FROM sys.ints AS t
JOIN sys.ints AS u
JOIN sys.ints AS v
WHERE ( t.i*100 + u.i*10 + v.i ) < 1000
ORDER BY Date;
This can be done down to the level of seconds:
drop table if exists secs;
create table secs
select sec_to_time( t.i*10000 + u.i*1000 + v.i*100 + w.i*10 + x.i ) as sec
from sys.ints t
join sys.ints u
join sys.ints v
join sys.ints w
join sys.ints x
where sec_to_time( t.i*10000 + u.i*1000 + v.i*100 + w.i*10 + x.i ) < '24:00:00'
order by sec;
You can use such a query as a View, as an inline derived table, or as input to a
CREATE TABLE statement.
A slightly more elaborate method, giving the calendar table an auto-increment key that can also be used as a surrogate for datetime interval calculations:
CREATE TABLE calendar (
id INT AUTO_INCREMENT PRIMARY KEY,
date DATE,
UNIQUE days (date)
);
Calculate the number of days needed in the calendar, eg
SELECT DATEDIFF('2010-12-31','1989-12-31'); # 7670, or 21*365 plus 5
Find a table with that many rows, 7670 in this case. Add a row to the calendar table for every day in the range:
INSERT INTO calendar (id)
SELECT NULL FROM [name of table with 7670 rows] LIMIT 4018;
Populate the date column by incrementing the starting date:
UPDATE calendar SET date = ADDDATE('1989-12-31',id);
The calendar table now has one row for each day from 1990-01-01 through 2010-12-31. Keep the auto_increment ID column for quick day counts in the range, or drop the column if you don't need that.
To make the calendar table a diary, make the period one leap year, add month, day and text columns, update month and day values with
MONTH(date) and
DAYOFMONTH(date) respectively, and if the diary is to be used from year to year, drop the date field.
[Based on a
builder.com SQL Tip by Arthur Fuller and a MySQL list tip by Michael Stassen]
To automate all this, write a stored procedure, for example:
CREATE TABLE times (
date_hour DATETIME,
KEY ( date_hour )
);
DROP PROCEDURE IF EXISTS timespopulate;
DELIMITER |
CREATE PROCEDURE timespopulate( startdate DATETIME, num INT )
BEGIN
DECLARE ctr INT DEFAULT 0;
WHILE ctr < num DO
BEGIN
INSERT INTO times VALUES ( DATE_ADD( startdate, INTERVAL ctr HOUR) );
SET ctr = ctr + 1;
END;
END WHILE;
END;
|
DELIMITER ;
CALL timespopulate( '2007-1-1, 31*24 );
Or, you can have the sproc do your counting:
DROP PROCEDURE IF EXISTS calendar;
DELIMITER |
CREATE PROCEDURE calendar( pstart datetime, pstop datetime, pminutes int )
DETERMINISTIC
BEGIN
DECLARE thisdate datetime;
DROP TABLE IF EXISTS cal;
CREATE TABLE cal( dt datetime );
SET thisdate=pstart;
INSERT INTO cal VALUES(pstart);
WHILE thisdate < pstop DO
SET thisdate = adddate( thisdate, INTERVAL pminutes MINUTE );
INSERT INTO cal VALUES( thisdate );
END WHILE;
END |
DELIMITER ;
-- make cal for 2007, 20-min intervals:
CALL calendar('2007-1-1 00:00:00', '2007-2-1 00:00:00', 20);
Peak visit counts by datetime period
You have a visits table (id int, start datetime, end datetime), and you wish to track peak visit counts. A simple solution is to self-join on non-matching IDs and overlapping visit times, group by ID, then order by the resulting counts:
SELECT a.id,group_concat(b.id) as Overlaps, count(b.id)+1 as OverlapCount
FROM visits a
JOIN visits b on a.id < b.id and a.start < b.end and b.start < a.end
GROUP BY a.id
ORDER BY OverlapCount DESC;
Scope to the week of a given date
To scope a query to the calendar week (Sunday through Saturday) of a date value @d, write ...
... WHERE d BETWEEN AddDate(@d,-DayOfWeek(@d)+1) and AddDate(@d,7-DayOfWeek(@d)) ...
Last updated 29 Dec 2009
Sum accumulated time by date
You track resource booking periods. You need a query to report daily usage for a given resource.
First the problem of calculating per-diem usage. Call the starting datetime of a booked period pStart, and its ending datetime pEnd. Then for a given date pDate, if the period began before pDate. then pDate usage begins at 00:00:00, otherwise it starts at pStart; likewise if the period extends past pDate, then pDate usage ends at midnight on pDate, otherwise it ends at pEnd. Therefore the period begins at...
IF( pStart < pDate, CAST(pDate AS DATETIME ), pStart )
and ends at...
IF( DATE(pEnd) > pDate, CAST(ADDDATE(pDate, 1) AS DATETIME), pEnd )
so pDate usage in seconds is given by...
UNIX_TIMESTAMP( IF( DATE(pEnd) > pDate, CAST(ADDDATE(pDate, 1) AS DATETIME), pEnd )) -
UNIX_TIMESTAMP( IF( pStart < pDate, CAST(pDate AS DATETIME ), pStart ));
To help keep the query as uncluttered as possible, encapsulate the above calculation in a stored function...
SET GLOBAL log_bin_trust_function_creators=1;
DROP FUNCTION IF EXISTS DaySeconds;
CREATE FUNCTION DaySeconds( pStart datetime, pEnd datetime, pDate date ) RETURNS INT
RETURN UNIX_TIMESTAMP( IF( DATE(pEnd) > pDate, CAST(ADDDATE(pDate, 1) AS DATETIME), pEnd )) -
UNIX_TIMESTAMP( IF( pStart < pDate, CAST(pDate AS DATETIME ), pStart ));
SELECT CEIL(DaySeconds('2008-1-1 10:05:00','2008-1-1 10:59:30','2008-1-1')/60) AS Mins;
+------+
| Mins |
+------+
| 55 |
+------+
Now for the query. To report usage per day over a period, we need a calendar table. For a test, make one for the first 100 days of the year 2008:
DROP TABLE IF EXISTS ints,calendar;
CREATE TABLE INTS(i int);
INSERT INTO ints VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
CREATE TABLE calendar(date date);
SET @n=0;
INSERT INTO calendar SELECT ADDDATE('2008-1-1',@n:=@n+1) FROM sys.ints a JOIN sys.ints b;
Now some test bookings data:
CREATE TABLE bookings( id INT PRIMARY KEY, resourceID int, startdate datetime, enddate datetime );
INSERT INTO bookings VALUES
(1,1,'2008-02-03 17:05','2008-02-03 19:00'),
(2,1,'2008-02-04 17:05','2008-02-04 18:00'),
(3,1,'2008-02-04 19:30','2008-02-04 20:00'),
(4,1,'2008-02-05 23:05','2008-02-06 01:00'),
(5,2,'2008-02-05 14:05','2008-02-05 15:00');
Obtain usage per day for resourceID=1 by running DaySeconds() against a JOIN of calendar and bookings:
SELECT
c.date AS date,
SUM( CEIL(( DaySeconds( b.startdate, b.enddate, c.date ) / 60 ))) AS 'Mins Used'
FROM calendar c
JOIN bookings b ON c.date BETWEEN DATE(b.startdate) AND DATE(b.enddate)
WHERE b.resourceID = 1
GROUP BY c.date;
+------------+-----------+
| date | Mins Used |
+------------+-----------+
| 2008-02-03 | 115 |
| 2008-02-04 | 85 |
| 2008-02-05 | 55 |
| 2008-02-06 | 60 |
+------------+-----------+
To report over a date range, join the above to the calendar table on a date range:
SELECT c.date, IFNULL( sums.N, 0 ) AS 'Mins Used'
FROM calendar AS c
LEFT JOIN (
SELECT
c.date AS date,
SUM( CEIL(( DaySeconds( b.startdate, b.enddate, c.date ) / 60 ))) AS N
FROM calendar c
JOIN bookings b ON c.date BETWEEN DATE(b.startdate) AND DATE(b.enddate)
WHERE b.resourceID = 1
GROUP BY c.date
) AS sums ON sums.date = c.date
WHERE c.date BETWEEN '2008-02-01' AND '2008-02-10';
+------------+-----------+
| date | Mins Used |
+------------+-----------+
| 2008-02-01 | 0 |
| 2008-02-02 | 0 |
| 2008-02-03 | 115 |
| 2008-02-04 | 85 |
| 2008-02-05 | 55 |
| 2008-02-06 | 60 |
| 2008-02-07 | 0 |
| 2008-02-08 | 0 |
| 2008-02-09 | 0 |
| 2008-02-10 | 0 |
+------------+-----------+
{based on a contribution by Daniel Patterson]
Sum for time periods
A table tracks attendance at some location:
drop table if exists t;
create table t(interval_id int,start datetime,end datetime, att int);
insert into t values
(1,'2007-01-01 08:00:00','2007-01-01 12:00:00',5 ),
(2,'2007-01-01 13:00:00','2007-01-01 17:00:00',10),
(3,'2007-01-01 10:00:00','2007-01-01 15:00:00',15),
(4,'2007-01-01 14:00:00','2007-03-07 19:00:00',20);
select * from t;
+-------------+---------------------+---------------------+------+
| interval_id | start | end | att |
+-------------+---------------------+---------------------+------+
| 1 | 2007-01-01 08:00:00 | 2007-01-01 12:00:00 | 5 |
| 2 | 2007-01-01 13:00:00 | 2007-01-01 17:00:00 | 10 |
| 3 | 2007-01-01 10:00:00 | 2007-01-01 15:00:00 | 15 |
| 4 | 2007-01-01 14:00:00 | 2007-03-07 19:00:00 | 20 |
+-------------+---------------------+---------------------+------+
In this table, att is a delta: it tracks entrances. Actual attendance is SUM(att) at any given moment. For example, if the attendance sum is x at a given moment, then after we add a row with att=y and start/end datetimes embracing that moment, attendance will be x+y. So to retrieve total attendance at 01330h on 1 Jan 2007, we write:
SELECT SUM(att)
FROM t
WHERE t.start <= '2007-01-01 13:30:00' AND t.end >= '2007-01-01 13:30:00';
+----------+
| SUM(att) |
+----------+
| 25 |
+----------+
Then how would we extract maximum attendance during a given period, for example, maximum attendance between 1300h and 1700h?
SQL does not deal efficiently with time. Some SQL dialects offer time series enhancements to the language; MySQL does not.
And, querying time series data for aggregate statistics gets complicated very quickly.
It gets a bit simpler with a calendar table that has a row for every possible datetime value. For our example, assume a granularity of one hour and a query period of one day. Naturally a real system would require a range of dates and perhaps a finer time granularity:
create table cal(id int,dt datetime);
insert into cal values(1,'2007-1-1 01:00:00');
insert into cal values(2,'2007-1-1 02:00:00');
insert into cal values(3,'2007-1-1 03:00:00');
insert into cal values(4,'2007-1-1 04:00:00');
insert into cal values(5,'2007-1-1 05:00:00');
insert into cal values(6,'2007-1-1 06:00:00');
insert into cal values(7,'2007-1-1 07:00:00');
insert into cal values(8,'2007-1-1 08:00:00');
insert into cal values(9,'2007-1-1 09:00:00');
insert into cal values(10,'2007-1-1 10:00:00');
insert into cal values(11,'2007-1-1 11:00:00');
insert into cal values(12,'2007-1-1 12:00:00');
insert into cal values(13,'2007-1-1 13:00:00');
insert into cal values(14,'2007-1-1 14:00:00');
insert into cal values(15,'2007-1-1 15:00:00');
insert into cal values(16,'2007-1-1 16:00:00');
insert into cal values(17,'2007-1-1 17:00:00');
insert into cal values(18,'2007-1-1 18:00:00');
insert into cal values(19,'2007-1-1 19:00:00');
insert into cal values(20,'2007-1-1 20:00:00');
insert into cal values(21,'2007-1-1 21:00:00');
insert into cal values(22,'2007-1-1 22:00:00');
insert into cal values(23,'2007-1-1 23:00:00');
insert into cal values(24,'2007-1-1 24:00:00');
To accumulate the maximum attendance sum, collect target values for defined periods in an inner query, and sum them from the outer query:
SELECT SUM( att )
FROM (
SELECT
t.start AS PeriodStart,
t.end AS PeriodEnd,
MIN(cal.dt) + INTERVAL 1 HOUR AS CountBegin,
MAX(cal.dt) AS CountEnd,
t.att
FROM t
JOIN cal ON cal.dt >= t.start AND cal.dt < t.end
GROUP BY PeriodStart, PeriodEnd
HAVING CountBegin < '2007-01-01 17:00:00' AND CountEnd > '2007-01-01 11:00:00'
) AS periods;
+------------+
| SUM( att ) |
+------------+
| 45 |
+------------+
If the data is more complicated, eg if we also need to track exits, the period logic needs refinement but the principle remains the same.
Sum time values
SELECT SEC_TO_TIME( SUM( TIME_TO_SEC( time_col ))) AS total_time
FROM tbl;
Summing values like '12:65:23' produces meaningless results.
The date of next Thursday
Given a date and its weekday number (1=Sunday, ..., 7=Saturday), there are three possibilities:
1. Today is Thursday: then next Thursday is 7 days from now.
2. Today is before Thursday: then next Thursday is (5 minus today's weekday number) from now.
3. Today is after Thursday: then next Thursday is 7 + (5 minus today's weekday number).
set @d=curdate();
set @n = dayofweek(curdate());
select
@d:=adddate(curdate(),0) as date,
@n:=dayofweek(adddate(curdate(),0)) as weekday,
adddate(@d,if(@n=5,7,if(@n<5,5-@n,7+5-@n))) as thurs;
+------------+---------+------------+
| date | weekday | nextthurs |
+------------+---------+------------+
| 2008-03-10 | 2 | 2008-03-13 |
+------------+---------+------------+
It's easily encapsulated in a function:
set global log_bin_trust_function_creators=1;
drop function if exists dateofnextweekday;
delimiter |
create function dateofnextweekday( d date, which tinyint ) returns date
begin
declare today tinyint;
set today = dayofweek(d);
return adddate( d, if( today=which,7,if(today<which,which-today,7+which-today)));
end |
delimiter ;
While we're at it, we might as well have a function to return the most recent given weekday for a given date ...
delimiter go
create function dateoflastweekday( d date, which tinyint ) returns date
begin
declare today tinyint;
set today = dayofweek(d);
return adddate( d, if(today=which,-7,if(today>which,which-today,which-today-7) ));
end;
go
delimiter ;
Last updated 02 Dec 2009
Track when a value changed
You have a table that tracks a value and the time when the value was measured ...
drop table if exists changes;
create table changes(time time,value int);
insert into changes values
('00:00', 0 ),
('01:05', 1 ),
('01:09', 1 ),
('01:45', 1 ),
('02:24', 0 ),
('12:20', 1 ),
('12:40', 0 ),
('14:32', 0 ),
('18:20', 1 ),
('18:21', 1 ),
('20:40', 0 );
and you need a query to show the rows where the value changes.
For how to retrieve a rank ordering we can subsequently use, see
Rank order. Here we need rank ordering on time. We get it by joining each row R to rows with equal or earlier times and counting those rows:
DROP TABLE IF EXISTS tmp;
CREATE TABLE tmp
SELECT a.*, COUNT(a.value) AS rank
FROM changes a
JOIN changes b ON a.time >= b.time
GROUP BY a.time;
SELECT * FROM tmp;
+----------+-------+------+
| time | value | rank |
+----------+-------+------+
| 00:00:00 | 0 | 1 |
| 01:05:00 | 1 | 2 |
| 01:09:00 | 1 | 3 |
| 01:45:00 | 1 | 4 |
| 02:24:00 | 0 | 5 |
| 12:20:00 | 1 | 6 |
| 12:40:00 | 0 | 7 |
| 14:32:00 | 0 | 8 |
| 18:20:00 | 1 | 9 |
| 18:21:00 | 1 | 10 |
| 20:40:00 | 0 | 11 |
+----------+-------+------+
Now if we left join this result to itself on matching values and the left rank >= the right rank, and if we exclude rows with NULLs on the right side, we are left with just the rows for which the next value is different:
SELECT a.time, a.value
FROM tmp a
LEFT JOIN tmp b ON a.value=b.value AND a.rank=b.rank+1
WHERE b.rank IS NULL;
+----------+-------+
| time | value |
+----------+-------+
| 00:00:00 | 0 |
| 01:05:00 | 1 |
| 02:24:00 | 0 |
| 12:20:00 | 1 |
| 12:40:00 | 0 |
| 18:20:00 | 1 |
| 20:40:00 | 0 |
+----------+-------+
On MySQL 4.1 or later, this can all be done in one step:
SELECT a.time, a.value
FROM (
SELECT t1.*, COUNT(*) AS rank
FROM changes t1
LEFT JOIN changes t2 ON t1.time >= t2.time
GROUP BY t1.time
) AS a
LEFT JOIN (
SELECT t1.*, COUNT(*) AS rank
FROM changes t1
LEFT JOIN changes t2 ON t1.time >= t2.time
GROUP BY t1.time
) AS b ON a.rank = b.rank+1 AND a.value = b.value
WHERE b.time IS NULL
ORDER BY a.time;
(Every week or so, someone on a MySQL forum poses this question, and a forum regular who goes by the name 'laptop alias' posts a solution very like the above.)
What month does a week fall in?
"The" month of a week is ambiguous if the week straddles months. If we adopt the convention that the month of a week is the month of its beginning Sunday, then on 29 Nov 2009 ...
SET @weekno = Month CurDate() );
SET @date = AddDate('2009-01-01', 7*@weekno );
SET @day = DayOfWeek( @date );
SET @datecomp = IF( @day = 1, @date, AddDate( @date, 1-@day ));
SELECT @date,@day,@datecomp,Month(@datecomp) AS month;
+------------+------+------------+-------+
| @date | @day | @datecomp | month |
+------------+------+------------+-------+
| 2009-12-03 | 5 | 2009-11-29 | 11 |
+------------+------+------------+-------+
It's easy to encapsulate into a stored function. We might as well parameterise the daynumber which the function is to use as a criterion:
DROP FUNCTION IF EXISTS MonthOfWeek;
SET GLOBAL log_bin_trust_function_creators=1;
DELIMITER go
CREATE FUNCTION MonthOfWeek( pyear YEAR, pweek SMALLINT, pday SMALLINT ) RETURNS SMALLINT
BEGIN
DECLARE vdate DATE;
DECLARE vday SMALLINT;
SET vdate = AddDate(Concat( pyear, '-01-01'), 7*pweek );
SET vday = DayOfWeek( vdate );
SET vdate = IF( vday = 1, vdate, AddDate( vdate, 1-vday ));
RETURN Month( vdate );
END;
go
DELIMITER ;
SELECT MonthOfWeek( Year(CurDate()), Week( CurDate() ), 1 );
Last updated 30 Nov 2009
Game schedule
List all possible home-away encounters of teams listed in a table.
SELECT t1.name AS Visiting,
t2.name AS Home
FROM teams AS t1
STRAIGHT_JOIN teams AS t2
WHERE t1.ID <> t2.ID;
Pivot table schedule
You have a schedule table (period, day, subject, room) with a primary key period,day to avoid duplicate bookings. You wish to display the schedule as periods, subjects and rooms in rows, and days of the week in columns.
SELECT
period,
MAX(IF(day=1, CONCAT(subject,' ',room), '')) AS Mon,
MAX(IF(day=2, CONCAT(subject,' ',room), '')) AS Tue,
MAX(IF(day=3, CONCAT(subject,' ',room), '')) AS Wed,
MAX(IF(day=4, CONCAT(subject,' ',room), '')) AS Thu,
MAX(IF(day=5, CONCAT(subject,' ',room), '')) AS Fri
FROM schedule
GROUP BY period
MAX() chooses existing over blank entries, and GROUP BY lines everything up on the same row.
Display column values which occur N times
SELECT id
FROM tbl
GROUP BY id
HAVING COUNT(*) = N;
Change the HAVING condition to >1 to list duplicate values, etc.
Display every Nth row
Display every Nth row in tbl where id>id is sequential in MySQL before version 4.1:
SELECT id
FROM tbl
GROUP BY id
HAVING MOD(id, N) = 0;
or
SELECT a.*
FROM tbl a
JOIN tbl b ON a.id = b.id AND MOD(b.id, N ) = 0;
In MySQL 4.1 and later you can also write:
SELECT *
FROM tbl
WHERE ( id, 0 ) IN ( SELECT id, MOD( id, N ) FROM tbl );
or
SELECT *
FROM ( SELECT id FROM tbl ) AS tmp
WHERE MOD( tmp.id, N ) = 0;
Last updated 14 Feb 2010
Trees, networks and parts explosions in MySQL
http://www.artfulsoftware.com/mysqlbook/sampler/mysqled1ch20.html
Dijkstra’s shortest path algorithm
Given a table of source-to-destination paths, each of whose nodes references a row in a nodes table, how do we find the shortest path from one node to another?
One answer is Dijkstra's algorithm (
http://en.wikipedia.org/wiki/Dijkstra's_algorithm). Peter Larsson has posted a SQL Server implementation of it on the
SQL Team Forum. Here is a MySQL implementation.
The DDL:
DROP TABLE IF EXISTS dijnodes,dijpaths;
CREATE TABLE dijnodes (
nodeID int PRIMARY KEY AUTO_INCREMENT NOT NULL,
nodename varchar (20) NOT NULL,
cost int NULL,
pathID int NULL,
calculated tinyint NOT NULL
);
CREATE TABLE dijpaths (
pathID int PRIMARY KEY AUTO_INCREMENT,
fromNodeID int NOT NULL ,
toNodeID int NOT NULL ,
cost int NOT NULL
);
Here is a stored procedure to populate valid nodes and paths:
DROP PROCEDURE IF EXISTS dijAddPath;
DELIMITER |
CREATE PROCEDURE dijAddPath(
pFromNodeName VARCHAR(20), pToNodeName VARCHAR(20), pCost INT
)
BEGIN
DECLARE vFromNodeID, vToNodeID, vPathID INT;
SET vFromNodeID = ( SELECT NodeID FROM dijnodes WHERE NodeName = pFromNodeName );
IF vFromNodeID IS NULL THEN
BEGIN
INSERT INTO dijnodes (NodeName,Calculated) VALUES (pFromNodeName,0);
SET vFromNodeID = LAST_INSERT_ID();
END;
END IF;
SET vToNodeID = ( SELECT NodeID FROM dijnodes WHERE NodeName = pToNodeName );
IF vToNodeID IS NULL THEN
BEGIN
INSERT INTO dijnodes(NodeName, Calculated)
VALUES(pToNodeName,0);
SET vToNodeID = LAST_INSERT_ID();
END;
END IF;
SET vPathID = ( SELECT PathID FROM dijpaths
WHERE FromNodeID = vFromNodeID AND ToNodeID = vToNodeID
);
IF vPathID IS NULL THEN
INSERT INTO dijpaths(FromNodeID,ToNodeID,Cost)
VALUES(vFromNodeID,vToNodeID,pCost);
ELSE
UPDATE dijpaths SET Cost = pCost
WHERE FromNodeID = vFromNodeID AND ToNodeID = vToNodeID;
END IF;
END;
|
DELIMITER ;
Use dijAddpath() to populate the tables:
call dijaddpath( 'a', 'b', 4 );
call dijaddpath( 'a', 'd', 1 );
call dijaddpath( 'b', 'a', 74 );
call dijaddpath( 'b', 'c', 2 );
call dijaddpath( 'b', 'e', 12 );
call dijaddpath( 'c', 'b', 12 );
call dijaddpath( 'c', 'f', 74 );
call dijaddpath( 'c', 'j', 12 );
call dijaddpath( 'd', 'e', 32 );
call dijaddpath( 'd', 'g', 22 );
call dijaddpath( 'e', 'd', 66 );
call dijaddpath( 'e', 'f', 76 );
call dijaddpath( 'e', 'h', 33 );
call dijaddpath( 'f', 'i', 11 );
call dijaddpath( 'f', 'j', 21 );
call dijaddpath( 'g', 'd', 12 );
call dijaddpath( 'g', 'h', 10 );
call dijaddpath( 'h', 'g', 2 );
call dijaddpath( 'h', 'i', 72 );
call dijaddpath( 'i', 'f', 31 );
call dijaddpath( 'i', 'j', 7 );
call dijaddpath( 'i', 'h', 18 );
call dijaddpath( 'j', 'f', 8 );
SELECT * FROM dijnodes;
+--------+----------+------+--------+------------+
| nodeID | nodename | cost | pathID | calculated |
+--------+----------+------+--------+------------+
| 1 | a | NULL | NULL | 0 |
| 2 | b | NULL | NULL | 0 |
| 3 | d | NULL | NULL | 0 |
| 4 | c | NULL | NULL | 0 |
| 5 | e | NULL | NULL | 0 |
| 6 | f | NULL | NULL | 0 |
| 7 | j | NULL | NULL | 0 |
| 8 | g | NULL | NULL | 0 |
| 9 | h | NULL | NULL | 0 |
| 10 | i | NULL | NULL | 0 |
+--------+----------+------+--------+------------+
SELECT * FROM dijpaths;
+--------+------------+----------+------+
| pathID | fromNodeID | toNodeID | cost |
+--------+------------+----------+------+
| 1 | 1 | 2 | 4 |
| 2 | 1 | 3 | 1 |
| 3 | 2 | 1 | 74 |
| 4 | 2 | 4 | 2 |
| 5 | 2 | 5 | 12 |
| 6 | 4 | 2 | 12 |
| 7 | 4 | 6 | 74 |
| 8 | 4 | 7 | 12 |
| 9 | 3 | 5 | 32 |
| 10 | 3 | 8 | 22 |
| 11 | 5 | 3 | 66 |
| 12 | 5 | 6 | 76 |
| 13 | 5 | 9 | 33 |
| 14 | 6 | 10 | 11 |
| 15 | 6 | 7 | 21 |
| 16 | 8 | 3 | 12 |
| 17 | 8 | 9 | 10 |
| 18 | 9 | 8 | 2 |
| 19 | 9 | 10 | 72 |
| 20 | 10 | 6 | 31 |
| 21 | 10 | 7 | 7 |
| 22 | 10 | 9 | 18 |
| 23 | 7 | 6 | 8 |
+--------+------------+----------+------+
Now for the stored procedure, a 6-step:
- null out path columns in the nodes table
- find the nodeIDs referenced by input params
- loop through all uncalculated one-step paths, calculating costs in each
- if a node remains uncalculated, the graph is invalid, so quit
- write the path sequence to a temporary table
- query the temp table to show the result
DROP PROCEDURE IF EXISTS dijResolve;
DELIMITER |
CREATE PROCEDURE dijResolve( pFromNodeName VARCHAR(20), pToNodeName VARCHAR(20) )
BEGIN
DECLARE vFromNodeID, vToNodeID, vNodeID, vCost, vPathID INT;
DECLARE vFromNodeName, vToNodeName VARCHAR(20);
-- null out path info in the nodes table
UPDATE dijnodes SET PathID = NULL,Cost = NULL,Calculated = 0;
-- find nodeIDs referenced by input params
SET vFromNodeID = ( SELECT NodeID FROM dijnodes WHERE NodeName = pFromNodeName );
IF vFromNodeID IS NULL THEN
SELECT CONCAT('From node name ', pFromNodeName, ' not found.' );
ELSE
BEGIN
-- start at src node
SET vNodeID = vFromNodeID;
SET vToNodeID = ( SELECT NodeID FROM dijnodes WHERE NodeName = pToNodeName );
IF vToNodeID IS NULL THEN
SELECT CONCAT('From node name ', pToNodeName, ' not found.' );
ELSE
BEGIN
-- calculate path costs till all are done
UPDATE dijnodes SET Cost=0 WHERE NodeID = vFromNodeID;
WHILE vNodeID IS NOT NULL DO
BEGIN
UPDATE
dijnodes AS src
JOIN dijpaths AS paths ON paths.FromNodeID = src.NodeID
JOIN dijnodes AS dest ON dest.NodeID = Paths.ToNodeID
SET dest.Cost = CASE
WHEN dest.Cost IS NULL THEN src.Cost + Paths.Cost
WHEN src.Cost + Paths.Cost < dest.Cost THEN src.Cost + Paths.Cost
ELSE dest.Cost
END,
dest.PathID = Paths.PathID
WHERE
src.NodeID = vNodeID
AND (dest.Cost IS NULL OR src.Cost + Paths.Cost < dest.Cost)
AND dest.Calculated = 0;
UPDATE dijnodes SET Calculated = 1 WHERE NodeID = vNodeID;
SET vNodeID = ( SELECT nodeID FROM dijnodes
WHERE Calculated = 0 AND Cost IS NOT NULL
ORDER BY Cost LIMIT 1
);
END;
END WHILE;
END;
END IF;
END;
END IF;
IF EXISTS( SELECT 1 FROM dijnodes WHERE NodeID = vToNodeID AND Cost IS NULL ) THEN
-- problem, cannot proceed
SELECT CONCAT( 'Node ',vNodeID, ' missed.' );
ELSE
BEGIN
-- write itinerary to map table
DROP TEMPORARY TABLE IF EXISTS map;
CREATE TEMPORARY TABLE map (
RowID INT PRIMARY KEY AUTO_INCREMENT,
FromNodeName VARCHAR(20),
ToNodeName VARCHAR(20),
Cost INT
) ENGINE=MEMORY;
WHILE vFromNodeID <> vToNodeID DO
BEGIN
SELECT
src.NodeName,dest.NodeName,dest.Cost,dest.PathID
INTO vFromNodeName, vToNodeName, vCost, vPathID
FROM
dijnodes AS dest
JOIN dijpaths AS Paths ON Paths.PathID = dest.PathID
JOIN dijnodes AS src ON src.NodeID = Paths.FromNodeID
WHERE dest.NodeID = vToNodeID;
INSERT INTO Map(FromNodeName,ToNodeName,Cost) VALUES(vFromNodeName,vToNodeName,vCost);
SET vToNodeID = (SELECT FromNodeID FROM dijPaths WHERE PathID = vPathID);
END;
END WHILE;
SELECT FromNodeName,ToNodeName,Cost FROM Map ORDER BY RowID DESC;
DROP TEMPORARY TABLE Map;
END;
END IF;
END;
|
DELIMITER ;
CALL dijResolve( 'a','i');
+--------------+------------+------+
| FromNodeName | ToNodeName | Cost |
+--------------+------------+------+
| a | b | 4 |
| b | c | 6 |
| c | j | 18 |
| j | f | 26 |
| f | i | 37 |
+--------------+------------+------+
Trees of known depth
A tree is a hierarchy where each node except the root has one parent. A parent-child link is an
edge. Edges in a tree of known depth can be queried with n-1 self-joins where n is the number of edges from top to bottom. Here is a simple example of depth 2:
drop table if exists t;
create table t(id int, parent int, ord int, title char(20));
insert into t values
(1, 0, 0, 'Root'),
(2, 1, 0, 'Home'),
(3, 1, 1, 'Projects'),
(5, 1, 2, 'Secret area'),
(4, 1, 3, 'Tutorials'),
(8, 1, 4, 'Example'),
(6, 4, 0, 'Computing'),
(7, 4, 1, 'Life');
To list children under their parents, self-join on left parent = right id and order on left parent, right ord:
SELECT t1.id,t1.parent,t2.ord,t1.title,t1.title as parentTitle
FROM t t1
LEFT JOIN t t2 ON t1.parent=t2.id
ORDER BY t1.parent,t2.ord;
+------+--------+------+-------------+-------------+
| id | parent | ord | title | parentTitle |
+------+--------+------+-------------+-------------+
| 1 | 0 | NULL | Root | Root |
| 8 | 1 | 0 | Example | Example |
| 2 | 1 | 0 | Home | Home |
| 3 | 1 | 0 | Projects | Projects |
| 5 | 1 | 0 | Secret area | Secret area |
| 4 | 1 | 0 | Tutorials | Tutorials |
| 6 | 4 | 3 | Computing | Computing |
| 7 | 4 | 3 | Life | Life |
+------+--------+------+-------------+-------------+
For trees of unknown depth see
http://artfulsoftware.com/mysqlbook/sampler/mysqled1ch20.html
Approximate joins
There are two main ways to reconcile payments against charges:
- Open Item: match payments against individual charges, typically by carrying the charge number in the payments table
,
- Statement: list and sum all charges and all payments, and show the difference as the outstanding balance.
The Open Item method needs a foolproof way to match payments to charges, but what if the customer neglected to return a copy of the invoice, or to write the invoice number on the cheque? Reconciliation staff spend much of their time resolving such problems.
Can we help? Yes! It won't be entirely foolproof, but it will drastically cut down the onerous work of reconciliation.
Here is DDL for a test case:
CREATE SCHEMA approx;
USE approx;
CREATE TABLE charges (
ID INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
custID INT UNSIGNED,
amount DECIMAL(10,2) NOT NULL
);
CREATE TABLE payments (
ID INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
custID INT UNSIGNED,
amount DECIMAL( 10,2) NOT NULL
);
Both tables carry a
custID column to identify whose charge or payment it is, but there is no foreign key linking payments to specific charges--that is the link we are going to approximate.
Now populate the tables with a few rows of sample charges and payments for customer #1, ensuring that you have a variety of payments – some that match the charge exactly, some that are close but not enough, and some that are slight overpayments.
INSERT INTO approx.charges VALUES
(NULL,1,100),(NULL,1,12),(NULL,1,56),(NULL,1,43),(NULL,1,59),(NULL,1,998);
INSERT INTO approx.payments VALUES
(NULL,1,99),(NULL,1,62),(NULL,1,40),(NULL,1,50),(NULL,1,12),(NULL,1,1000);
SELECT * FROM charges;
+----+--------+--------+
| ID | custID | amount |
+----+--------+--------+
| 1 | 1 | 100.00 |
| 2 | 1 | 12.00 |
| 3 | 1 | 56.00 |
| 4 | 1 | 43.00 |
| 5 | 1 | 59.00 |
| 6 | 1 | 998.00 |
+----+--------+--------+
SELECT * FROM payments;
+----+--------+---------+
| ID | custID | amount |
+----+--------+---------+
| 1 | 1 | 99.00 |
| 2 | 1 | 62.00 |
| 3 | 1 | 40.00 |
| 4 | 1 | 50.00 |
| 5 | 1 | 12.00 |
| 6 | 1 | 1000.00 |
+----+--------+---------+
The first thing to do is define an approximation threshold: how close must the amount paid be to the amount charged before we conclude that the amounts are related? For this example we define the proximity threshold as 2. In a real-world example, it might be 10, or 50, or perhaps percentage of the charge. It all depends on the nature of the organisation and the typical total purchase. A house builder may make frequent purchases valued at $1000 and more. You scale the threshold to the typical situation.
Since the amount paid might be more or less or even equal to the amount charged, to link a payment to a charge we need not an
equi-join but a
theta-join that tests a range both below and above the charge amount. That might suggest a
BETWEEN clause. Here is a better idea: use the
ABS() function:
SET @proximity = 2; -- change this value to suit your situation
SELECT
c.ID AS ChargeNo,
c.Amount AS Charge,
p.ID AS PaymentNo,
p.Amount AS Payment
FROM charges c
JOIN payments p
ON c.custID = p.custID
AND ABS(c.amount - p.amount) <= @proximity
WHERE c.custID = 1;
Before you run this query, look at the data to anticipate the result.
Here it is:
+----------+--------+-----------+---------+
| ChargeNo | Charge | PaymentNo | Payment |
+----------+--------+-----------+---------+
| 1 | 100.00 | 1 | 99.00 |
| 2 | 12.00 | 5 | 12.00 |
| 6 | 998.00 | 6 | 1000.00 |
+----------+--------+-----------+---------+
The solution is correct, as far as it goes, but it doesn’t go far enough. We correctly identified the three situations: underpayment, exact payment and overpayment, but we suppressed all charges that don’t have a matching payment. Reconciliation staff are probably interested in a bigger picture of the situation. Fix this by changing the
INNER JOIN to a
LEFT JOIN:
SET @proximity = 2;
SELECT
c.ID AS ChargeNo,
c.amount AS Charge,
p.ID AS PaymentNo,
p.amount AS Payment
FROM
charges c
LEFT JOIN payments p
ON c.custID = p.custID
AND ABS(c.amount - p.amount) <= @proximity
WHERE c.custID = 1;
+----------+--------+-----------+---------+
| ChargeNo | Charge | PaymentNo | Payment |
+----------+--------+-----------+---------+
| 1 | 100.00 | 1 | 99.00 |
| 2 | 12.00 | 5 | 12.00 |
| 3 | 56.00 | NULL | NULL |
| 4 | 43.00 | NULL | NULL |
| 5 | 59.00 | NULL | NULL |
| 6 | 998.00 | 6 | 1000.00 |
+----------+--------+-----------+---------+
Much better! The reconciliation people now know that three charges have no matching payment.
What if the customer mistakenly pays for something twice? Add a row to the Payments table with an amount of $1000, then re-run the last query:
+----------+--------+-----------+---------+
| ChargeNo | Charge | PaymentNo | Payment |
+----------+--------+-----------+---------+
| 1 | 100.00 | 1 | 99.00 |
| 2 | 12.00 | 5 | 12.00 |
| 3 | 56.00 | NULL | NULL |
| 4 | 43.00 | NULL | NULL |
| 5 | 59.00 | NULL | NULL |
| 6 | 998.00 | 6 | 1000.00 |
| 6 | 998.00 | 7 | 1000.00 |
+----------+--------+-----------+---------+
How convenient! We can see at once that charge number 6 was paid for twice.
Somebody in the reconciliation department owes you lunch.
Cascading JOINs
Show parents, children and grandchildren including parents without children
SELECT parent.id AS ParentID,
IFNULL(child.parent_id,') AS ChildParentID,
IFNULL(child.id,') AS ChildID,
IFNULL(grandchild.child_id,') AS GrandchildChildID
FROM parent
LEFT JOIN child ON parent.id=child.parent_id
LEFT JOIN grandchild ON child.id=grandchild.child_id;
Data-driven joins
Data-driven table relationships are hard to maintain, but sometimes they cannot be avoided. How do we build joins for them? One way is to use a CASE statement in the SELECT list to handle the joining possibilities. In this example, the parent.linktable column determines the name of the table where a particular parent row's data is. The method is fine when the number of child tables is small:
USE test;
DROP TABLE IF EXISTS parent, child1, child2;
CREATE TABLE parent (
id INT UNSIGNED PRIMARY KEY,
linktable CHAR(64) NOT NULL
);
INSERT INTO parent VALUES (1, 'child1'), (2, 'child2');
CREATE TABLE child1 (
id INT UNSIGNED PRIMARY KEY,
data CHAR(10)
);
INSERT INTO child1 VALUES (1, 'abc');
CREATE TABLE child2 (
id INT UNSIGNED PRIMARY KEY,
data CHAR(10)
);
INSERT INTO child2 VALUES (2, 'def');
To retrieve all child data for all parents, include in the SELECT list a CASE statement which handles all child table possibilities:
SELECT
p.id,
p.linktable,
CASE linktable
WHEN 'child1' THEN c1.data
WHEN 'child2' THEN c2.data
ELSE 'Error'
END AS Data
FROM parent AS p
LEFT JOIN child1 AS c1 ON p.id=c1.id
LEFT JOIN child2 AS c2 ON p.id=c2.id;
+----+-----------+------+
| id | linktable | Data |
+----+-----------+------+
| 1 | child1 | abc |
| 2 | child2 | def |
+----+-----------+------+
When the number of child tables is too large for a convenient CASE statement, PREPARE the query in a stored procedure.
(Based on a MySQL Forum post by Felix Geerinckx)
Full Outer Join
A
FULL OUTER join between tables a and b retrieves:
- all rows from a, with matching rows or nulls from b, and
- all rows from b, with matching rows or nulls from a
so for these tables:
DROP TABLE IF EXISTS a,b;
CREATE TABLE a(id int,name char(1));
CREATE TABLE b(id int,name char(1));
INSERT INTO a VALUES(1,'a'),(2,'b');
INSERT INTO b VALUES(2,'b'),(3,'c');
SELECT * FROM a;
+------+------+
| id | name |
+------+------+
| 1 | a |
| 2 | b |
+------+------+
SELECT * FROM b;
+------+------+
| id | name |
+------+------+
| 2 | b |
| 3 | c |
+------+------+
a full outer join returns:
+------+------+------+------+
| id | name | id | name |
+------+------+------+------+
| 1 | a | NULL | NULL |
| 2 | b | 2 | b |
| NULL | NULL | 3 | c |
+------+------+------+------+
MySQL does not support
FULL OUTER JOIN. How to emulate it? If the joining keys of each table are unique, you can just
UNION left and right joins:
SELECT * FROM a LEFT JOIN b ON a.id=b.id
UNION
SELECT * FROM a RIGHT JOIN b ON a.id=b.id;
But suppose the tables to be joined have duplicate rows, and you wish your result to preserve them. For example, add a duplicate row to table a:
INSERT INTO a VALUES(1,'a');
Now
UNION removes the duplicate row you want preserved in the result. How to get back the desired duplicates? A
FULL OUTER JOIN consists of:
- an
INNER JOIN between a and b to catch row matches between a and b,
- a
LEFT EXCLUSION JOIN from a to b to catch rows that are in a and not in b,
- a
RIGHT EXCLUSION JOIN from b to a to catch rows in b that are not in a.
In SQL:
SELECT * FROM a INNER JOIN b ON a.id=b.id
UNION ALL
SELECT * FROM a LEFT JOIN b ON a.id=b.id WHERE b.id IS NULL
UNION ALL
SELECT * FROM a RIGHT JOIN b ON a.id=b.id WHERE a.id IS NULL
But the first two joins—the inner join, and the left exclusion join—are logically equivalent to a left outer join, so we can write:
SELECT * FROM a LEFT JOIN b ON a.id=b.id
UNION ALL
SELECT * FROM a RIGHT JOIN b ON a.id=b.id WHERE a.id IS NULL;
+------+------+------+------+
| id | name | id | name |
+------+------+------+------+
| 1 | a | NULL | NULL |
| 2 | b | 2 | b |
| 1 | a | NULL | NULL |
| NULL | NULL | 3 | c |
+------+------+------+------+
Why doesn't MySQL implement
FULL OUTER JOIN syntax for this? We don't know.
Intersection and difference
MySQL implements UNION, but does not directly implement INTERSECTION or DIFFERENCE.
INTERSECTION is just an INNER JOIN on all columns:
drop table if exists a,b;
create table a(i int,j int);
create table b like a;
insert into a values(1,1),(2,2);
insert into b values(1,1),(3,3);
select * from a join b using(i,j);
+------+------+
| i | j |
+------+------+
| 1 | 1 |
+------+------+
Get the DIFFERENCE between tables a and b by UNIONing exclusion joins from a to b, and from b to a:
select * from a left join b using(i,j) where b.i is null
union
select * from b left join a using(i,j) where a.i is null;
+------+------+
| i | j |
+------+------+
| 2 | 2 |
| 3 | 3 |
+------+------+
Last updated 30 Jun 2009
Many-to-many joins
To model a many:many relationship between two tables a and b, you need a bridging table where each row represents one instance of an association between a row in a and a row in b, as in this example:
drop table if exists users,actions,useractions;
create table users(userid int primary key, username char(32));
insert into users values(1, 'James'),(2, 'Alex'),(3, 'Justin');
create table actions(actionid int primary key, action char(32));
insert into actions values(1, 'Login'),(2, 'Logout'),(3, 'Delete'),(4, 'Promote');
create table useractions(uaid int primary key, userid int, actionid int);
insert into useractions values(1,1,1),(2,1,2),(3,3,4);
select u.username, a.action
from useractions ua
join users u using (userid)
join actions a using (actionid);
+----------+---------+
| username | action |
+----------+---------+
| James | Login |
| James | Logout |
| Justin | Promote |
+----------+---------+
Almost every non-trivial database will have at least one instance of this pattern. We have a collection of articles and users' scores of them. How to report statistics on these scores? We need three tables--one for articles, one for users, and a bridge table where each row represents one score on one article by one user:
DROP TABLE IF EXISTS art_articles;
CREATE TABLE art_articles (
ID INT AUTO_INCREMENT PRIMARY KEY,
title CHAR(30),
txt TEXT,
UNIQUE KEY (title)
);
INSERT INTO art_articles VALUES (1,'abc',''),(2,'def',''),(3,'ghi',''),(4,'jkl','');
DROP TABLE IF EXISTS art_users;
CREATE TABLE art_users(
ID INT AUTO_INCREMENT PRIMARY KEY,
name CHAR(20)
);
INSERT INTO art_users VALUES (1,'A'),(2,'B');
DROP TABLE IF EXISTS art_scores;
CREATE TABLE art_scores (
id INT AUTO_INCREMENT PRIMARY KEY,
articleID INT NOT NULL, -- references article.articleID
userID INT NOT NULL, -- references user.userID
score DECIMAL(6,2)
);
INSERT INTO art_scores VALUES (1,1,1,80),(2,1,2,90),(3,2,2,60);
-- find average score for article titled 'abc'
SELECT a.title, AVG( s.score ) AS Art1Avg
FROM art_articles a
JOIN art_scores s ON a.id=s.articleID
WHERE a.title='abc'
GROUP BY a.title;
-- find average score submitted by user 1
SELECT u.name, AVG( s.score ) AS User1Avg
FROM art_users u
JOIN art_scores s ON u.id=s.userID
WHERE u.id = 1
GROUP BY u.name;
Last updated 16 Mar 2010
What else did buyers of X buy?
We often want to know how certain column values associate with other column values, for example "What else did buyers of x buy?", or "What projects did Sean, Ian and Gerard all work on?"
Start with buyers of x. The table that summarises this information might be a View that encapsulates joins from customers to orders to orderitems to products, perhaps scoped on a recent date range. Here we ignore all such detail. We focus only on the logic of typical problems of this kind:
DROP TABLE IF EXISTS userpurchases;
CREATE TABLE userpurchases( custID INT UNSIGNED, prodID INT UNSIGNED );
INSERT INTO userpurchases
VALUES (1,1),(1,2),(2,4),(3,1),(3,2),(4,2),(4,3),(5,1),(5,2),(5,3);
SELECT custID, GROUP_CONCAT(prodID ORDER BY prodID) AS PurchaseList
FROM userpurchases
GROUP BY custID;
+--------+--------------+
| custID | PurchaseList |
+--------+--------------+
| 1 | 1,2 |
| 2 | 4 |
| 3 | 1,2 |
| 4 | 2,3 |
| 5 | 1,2,3 |
+--------+--------------+
The basic idea is to self-join on the prodID as often as necessary to get the answer. For example, to list all products bought by customers who'd already bought at least one other product, join userpurchases to itself on matching custIDs and non-matching prodIDs:
SELECT DISTINCT p2.prodid
FROM userpurchases p1
JOIN userpurchases p2 ON p1.custID = p2.custID AND p1.prodID <> p2.prodID;
+--------+
| prodid |
+--------+
| 1 |
| 2 |
| 3 |
+--------+
To find what else buyers of product 1 bought, copy the above join and group by custID:
SELECT p1.custID,GROUP_CONCAT(p2.prodid) as 'Buyers of #1 Also bought'
FROM userpurchases p1
JOIN userpurchases p2 ON p1.custID=p2.custID AND p1.prodID <> p2.prodID
WHERE p1.prodID = 1
GROUP BY p1.custID;
+--------+--------------------------+
| custID | Buyers of #1 Also bought |
+--------+--------------------------+
| 1 | 2 |
| 3 | 2 |
| 5 | 2,3 |
+--------+--------------------------+
What customers bought both product 1 and product 2?
SELECT DISTINCT p1.custID
FROM userpurchases p1
JOIN userpurchases p2 ON p1.custID=p2.custID AND p1.prodID=1 AND p2.prodID=2
+--------+
| custID |
+--------+
| 1 |
| 3 |
| 5 |
+--------+
Here is another solution: customers who bought products 1 and 2 ...
(i) have custIDs in userpurchases where prodID is in (1,2) and
(ii) have a distinct prodID count >= the number of prodIDs in (1,2), which is 2:
SELECT custID
FROM (
SELECT custID, COUNT(DISTINCT prodID) AS N
FROM userpurchases
WHERE prodID IN(1,2)
GROUP BY custID
HAVING N>=2 -- N of (1,2) = 2
) x;
When the criterion group has more than 2 items, that aggregating solution is easier to implement. For example, you track members and their projects, and you often have to query which members participated in all of a specified list of projects:
DROP TABLE IF EXISTS members,project_members;
CREATE TABLE members(emp_id int,emp_name char(5));
INSERT INTO members VALUES (1,'Emp1'),(2,'Emp2'), (3, Emp3'),(4,'Emp4'),(5,'Emp5');
CREATE TABLE project_members (project_id int, emp_id int);
INSERT INTO project_members
VALUES (1, 1),(1, 2),(1, 3),(2, 1),(2, 2),(3, 1),(4, 3),(5, 1),(5, 2),(5, 3),(5, 4),(6, 1);
SELECT project_id AS Project, GROUP_CONCAT(emp_id) AS Members
FROM project_members
GROUP BY project_id;
+---------+---------+
| Project | Members |
+---------+---------+
| 1 | 1,2,3 |
| 2 | 1,2 |
| 3 | 1 |
| 4 | 3 |
| 5 | 1,2,3,4 |
| 6 | 1 |
+---------+---------+
What projects did members 1, 2 and 3 work on?
SELECT project_id
FROM (
SELECT project_id, COUNT(DISTINCT emp_id) AS N
FROM project_members
WHERE emp_id IN(1,2,3)
GROUP BY project_id
HAVING N>=3 -- N of (1,2,3) = 3
) x;
+------------+
| project_id |
+------------+
| 1 |
| 5 |
+------------+
Here is a variation on the pattern. You have tables people(id, name), membership(id, name) and people_membership(id, membership_id, people_id). You need to find names of all persons and all their memberships where one of the memberships has id=X. It's another two-step:
1. Assemble the people who have a membership with id=X
2. Join from that derived result to people_memberships to get their other memberships, then to people to get their personal info:
SELECT
GROUP_CONCAT(m.name ORDER BY m.membership_id SEPARATOR ', ') AS 'Member Name'
pX.id,
pX.name
FROM (
SELECT p.id,p.id,p.name
FROM people p
JOIN (
SELECT id
FROM people_membership
WHERE membership_id=X
) pmX ON p.id=pmX.id
) pX
JOIN people_membership pm ON pX.id=pm.id
JOIN membership m ON pm.membership_id=m.membership_id
GROUP BY pX.id
ORDER BY pX.name;
Here is another variation on the pattern. You track companies, their agents, and their sales:
drop table if exists companies,agents,sales;
create table companies( id int,name char(5) );
insert into companies values (1,'a ltd'),(2,'b ltd');
create table agents( id int, agentname char(5), companyID int );
insert into agents values (1,'ann',1),(2,'bill',1),(3,'chad',2),(4,'david',2),(5,'ed',3);
create table sales( itemname char(5), agentID int );
insert into sales values('socks',1),('shoes',2),('shirt',3),('tie',4);
You need a report that lists all items sold by a given agent and by all other agents in that agent's company. It's a three-step:
1. Write a query to retrieve agents and their sales.
2. Join the result of [1] with the agents table on company.
3. Scope the result of [2] on a particular agent's name.
SET @agent='ann';
SELECT s.*
FROM agents a
JOIN (
SELECT x.id,x.agentname,x.companyID,y.itemname -- sales by agent
FROM agents x
JOIN sales y ON x.id=y.agentID
) s ON a.companyID=s.companyID -- users by company
WHERE a.agentname=@agent; -- scope by user
------+-----------+-----------+----------+
id | agentname | companyID | itemname |
------+-----------+-----------+----------+
1 | ann | 1 | socks |
2 | bill | 1 | shoes |
------+-----------+-----------+----------+
Last updated 15 Jan 2010
Join or subquery?
Usually, a JOIN is faster than an uncorrelated subquery. For example in the sakila test database, customer is a parent of rental (via customer_id) which in turn is a parent of payment (via rental_id). The subquery version of a query for whether a customer has made payments and rentals...
SELECT DISTINCT c.customer_id
FROM customer c
WHERE c.customer_id IN (
SELECT r.customer_id
FROM rental r
JOIN payment p USING (rental_id)
WHERE c.customer_id = 599;
);
is eight times slower than the join version...
SELECT DISTINCT c.customer_id
FROM customer c
JOIN rental r USING (customer_id)
JOIN payment p USING (rental_id)
WHERE c.customer_id = 599;
Running EXPLAIN on the two queries reveals why: the subquery version has to read most customer rows, while the join version proceeds inside out and discovers it needs to read just one customer row.
Parents without children
Given tables parent(id INT), child(id INT,parent_id INT), how do we find parents with no children? It's the All X for which there is no Y problem, which can be written as an exclusion join...
SELECT parent.id
FROM parent
LEFT JOIN child ON parent.id = child.parent_id
WHERE child.parent_id IS NULL;
or with a NOT EXISTS subquery, which is logically equivalent to the exclusion join, but usually performs much slower:
SELECT parent.id AS ParentID
FROM parent
WHERE NOT EXISTS (
SELECT parent.id
FROM parent
JOIN child ON parent.ID = child.parent_id
);
Parties who have contracts with one another
You have a parties table that holds info on peoples' names etc, and a contracts table where each row has clientID and contractorID value pointing at a parties.partyID value--that is, each contracts row points at two parties rows. You want to list the names of all contractors and their clients.
SELECT clientpartyID,
pCli.name AS Client,
contractorpartyID,
pCon.name AS Contractor
FROM contracts
INNER JOIN parties AS pCli
ON contracts.clientpartyID = pCli.partyID
INNER JOIN parties AS pCon
ON contracts.contractorpartyID = pCon.partyID;
The unbearable slowness of IN()
Intuitively,
IN() is easier to understand than
JOIN, so query beginners often write
IN() clauses instead of
JOINs, as for example in this query for
NorthWind orders having multiple items:
SELECT orderID
FROM orders
WHERE orderID IN (
SELECT orderID
FROM orderdetails
GROUP BY orderID
HAVING COUNT(orderID) > 1
);
IN() implements what is called a
semi-join; for each row from the left side of the join, it returns all matching rows from the right side. Unfortunately, without optimisations introduced in MySQL 6.0,
IN() queries can be impossibly slow. In MySQL 5.0 and 5.1, the above query is
130 times slower than its
JOIN version:
SELECT orderID
FROM orders o
JOIN orderdetails od USING (orderID)
GROUP BY orderID
HAVING COUNT(od.orderID) > 1
You wonder if
EXISTS() might be better?
SELECT orderID
FROM orders o
WHERE EXISTS (
SELECT orderID FROM orderdetails
WHERE orderID = o.orderID
GROUP BY orderID
HAVING COUNT(orderID)>1
);
In 5.0 and 5.1, it's twice as slow as the
JOIN version.
Why is the
JOIN version faster?
IN() and
EXISTS() subqueries have to execute a table scan for each row in the table. Performance degrades as the square of the number of rows. The
JOIN version builds its derived table on one table scan, and quickly picks off its resultset from that.
For
NOT IN(...), use a straight exclusion join. The basic pattern is that ...
SELECT ...
FROM a
WHERE a.x NOT IN (
SELECT b FROM y
);
becomes ...
SELECT ...
FROM a
LEFT JOIN b ON a.x=b.y
WHERE b.y IS NULL;
In MySQL 6.0, experiment with the new
optimizer_switch variable to see whether your semi-join runs faster with
IN(),
EXISTS() or
JOIN.
The [Not] Exists query pattern
Given a table
employee( employeeID INT, mgr_employeeID INT, salary DECIMAL(10,2)), find the managers who earn less than one or more of their subordinates.
We can write this query directly from the logic of its spec...
SELECT DISTINCT employeeID
FROM employee AS e
WHERE EXISTS (
SELECT employeeID
FROM employee AS m
WHERE m.mgr_employeeID = e.employeeID AND e.salary > m.salary
);
...but a
JOIN (or decorrelated) version of the logic is usually much faster. This query pattern is simple:
- Inner join the table (t1) to itself (t2) on the grouping key.
- Add the condition on which you wish to find existing rows to the Join clause.
SELECT DISTINCT m.employeeID
FROM employee AS e
INNER JOIN employee AS m ON e.Mgr = m.employeeID AND e.salary > m.salary;
The correlated subquery version of the
Not Exists query just inserts a strategic
NOT:
SELECT DISTINCT employeeID
FROM employee AS e
WHERE NOT EXISTS (
SELECT employeeID
FROM employee AS m
WHERE m.Mgr = e.employeeID AND e.salary > m.salary
);
The decorrelated version of
Not Exists uses an
exclusion join--a
LEFT JOIN with an
IS NULL condition imposed on the right side of the join:
- Left join the table to itself on the grouping key.
- Add the condition on which you wish to find existing rows to the Join clause.
- For the condition on which you wish to find missing rows on the right side,
- (a) add the value condition to the Join clause, and
- (b) in the Where clause, add an 'is null' condition for the right side:
SELECT DISTINCT m.employeeID
FROM employee AS e
LEFT JOIN employee AS m ON e.Mgr = m.employeeID AND e.salary > m.salary
WHERE m.employeeID IS NULL;
What exams did a student not register for?
We have a students table, an exams table, and a registrations table in which each row registers one student for one exam. How do we find the exams for which a particular student is not registered?
DROP TABLE IF EXISTS students, exams, registrations;
CREATE TABLE students (
sid int(10) unsigned PRIMARY KEY auto_increment,
firstname varchar(45) NOT NULL default '',
lastname varchar(45) NOT NULL default ''
);
INSERT INTO students VALUES
(1, 'Jack', 'Malone'),(2, 'Hiro', 'Nakamura'),(3, 'Bree', 'Van de Kamp'),
(4, 'Susan', 'Mayer'),(5, 'Matt', 'Parkman'),(6, 'Claire', 'Bennet');
CREATE TABLE exams (
eid int(10) unsigned PRIMARY KEY auto_increment,
exam_name varchar(45) NOT NULL default '',
active smallint(5) unsigned NOT NULL default '0'
);
INSERT INTO exams VALUES
(1, 'Javascript Expert', 1),(2, 'Lost Survival Course', 0),(3, 'Zend PHP Certification', 1),
(4, 'Superhero Advanced Skills', 1),(5, 'Desperation Certificate', 1);
CREATE TABLE registrations (
registration_id int(11) PRIMARY KEY auto_increment,
eid int(10) unsigned NOT NULL default '0',
sid int(10) unsigned NOT NULL default '0',
registration_date datetime NOT NULL default '0000-00-00 00:00:00'
);
INSERT INTO registrations (registration_id, eid, sid, registration_date) VALUES
(1, 5, 14, '2007-10-25 00:00:00'),(2, 5, 3, '0000-00-00 00:00:00'),
(3, 5, 4, '2007-10-23 00:00:00'),(4, 4, 2, '2007-10-16 00:00:00'),
(5, 4, 5, '2007-10-22 00:00:00'),(6, 4, 6, '2007-10-23 00:00:00'),
(7, 5, 2, '2007-10-23 00:00:00');
This is a version of the All X for which there is no Y problem, easily solved with a Not Exists query. But Not Exists queries can be painfully slow. Exclusion join solutions usually perform better, but here there is a wrinkle (thanks to Pascal Mitride for this example): we might expect to join registrations to students to get student info into the result, yet the registrations table will be the object of the exclusion join, so how do we retrieve the required student info?
A solution is to left join exams to a subquery: students left join registrations, restricted by a WHERE clause specifying the target student. The subquery encapsulates conditions on the joined table. Then we can impose the IS NULL condition on the subquery:
SELECT e.exam_name FROM exams AS e -- all exams
LEFT JOIN (
SELECT eid -- Hiro's exams
FROM students s LEFT JOIN registrations r ON s.sid=r.sid
WHERE s.firstname='Hiro' AND s.lastname='Nakamura'
) AS tmp ON e.eid=tmp.eid
WHERE tmp.eid IS NULL; -- exclusion
+------------------------+
| exam_name |
+------------------------+
| Javascript Expert |
| Lost Survival Course |
| Zend PHP Certification |
+------------------------+
A query showing Hiro's registration or not for all exams proves our logic correct:
SELECT e.exam_name,IF(tmp.sid IS NULL, 'No', 'Yes') AS 'Hiro registered'
FROM exams e
LEFT JOIN (
SELECT eid
FROM students s LEFT JOIN registrations r ON s.sid=r.sid
WHERE s.firstname='Hiro' AND s.lastname='Nakamura'
) tmp ON e.eid=tmp.eid;
+---------------------------+-----------------+
| exam_name | Hiro registered |
+---------------------------+-----------------+
| Javascript Expert | No |
| Lost Survival Course | No |
| Zend PHP Certification | No |
| Superhero Advanced Skills | Yes |
| Desperation Certificate | Yes |
+---------------------------+-----------------+
Beginners often have trouble working out how to write this kind of query, especially when the LEFT JOIN condition requires comparison with a literal value on the right side, thereby making the LEFT JOIN, effectively, an INNER JOIN and undermining the query design.
The method is to subtract one set (here, the exams Hiro registered for) from another (all exams). Start by writing a query expression for the set to be subtracted, then write the query expression to be subtracted from, then LEFT JOIN the two, then add the IS NULL condition:
1. Write a query expression to retrieve rows which positively match the exclusion condition. In simple cases, this may be a simple table reference. Here we need it to be an inner query for the exams Hiro Nakamura did register for...
SELECT eid
FROM students s LEFT JOIN registrations r ON s.sid=r.sid
WHERE s.firstname='Hiro' AND s.lastname='Nakamura'
2. Write a query to retrieve the rows from which the result of [1] is to be subtracted--here, simply all exams...
SELECT e.exam_name FROM exams e
3. Left join [2] to [1] on the appropriate key...
SELECT e.exam_name FROM exams AS e
LEFT JOIN (
SELECT eid
FROM students s LEFT JOIN registrations r ON s.sid=r.sid
WHERE s.firstname='Hiro' AND s.lastname='Nakamura'
) AS tmp ON e.eid=tmp.eid
4. Add a WHERE clause condition requiring that some column in the inner query evaluates to NULL.
WHERE tmp.eid IS NULL;
List NULLs at end of query output
If ordering by col...
... ORDER BY IF(col IS NULL, 0, 1 ), col ...
Last updated 30 Dec 2009
Parents with and without children
You have parties and contracts tables. Every contracts row has a contractorpartyID value which references a row in parties, and a clientpartyID value which also references a row in parties. How to list all parties and their contracts, showing blanks as empty strings rather than NULLs?
SELECT parties.partyID,
IFNULL(contractorpartyID,'') AS contractor,
IFNULL(clientpartyID,'') AS client
FROM parties
LEFT JOIN contractor_client ON partyID=contractorpartyID
ORDER BY partyID;
+---------+------------+--------+
| partyID | contractor | client |
+---------+------------+--------+
| 1 | | |
| 2 | 2 | 1 |
| 3 | | |
+---------+------------+--------+
Next row
You have a table of names, you have retrieved a row with name $name, and you want the row for the next name in name order. MySQL LIMIT syntax makes this very easy:
SELECT *
FROM tbl
WHERE name > $name
ORDER BY name
LIMIT 1
Order by leading digits
To have column values 1abc,10abc,8abc appear in the expected order 1abc,8abc,10abc, take advantage of a trick built into MySQL string parsing ...
SELECT '3xyz'+0;
+----------+
| '3xyz'+0 |
+----------+
| 3 |
+----------+
to write ...
SELECT ...
...
ORDER BY colname+0, colname;
Order by month name
The MySQL FIELD(str,str1,str2,...,strN) function returns 1 if str=str1, 2 if str=str2, etc., so ...
SELECT .
ORDER BY FIELD(month,'JAN','FEB','MAR',...,'NOV','DEC') .
will order query output from a legacy table in month-number order.
Order by numerics then alphas
Given a column type with values 1,2,3,a,b,c, how to get the order 3,2,1,c,b,a?
ORDER BY type RLIKE '^[0-9]+$' DESC, `type` DESC
Suppress repeating ordering values
You have tables tracking authors and their books, for example:
CREATE TABLE author (
id int(4) NOT NULL auto_increment PRIMARY KEY,
name text NOT NULL
);
INSERT INTO author (id, name)
VALUES (1,'Brad Phillips'),(2,'Don Charles'),(3,'Kur Silver');
CREATE TABLE book (
id int(4) NOT NULL auto_increment PRIMARY KEY,
name text NOT NULL
);
INSERT INTO book (id, name)
VALUES (1,'MySQL in a bucket '),(2,'Databases for Delinquents'),
(3,'Design Patterns'),(4,'PHP Professional'),(5,'Java Script Programming');
CREATE TABLE book_author (
book_id int(4) NOT NULL default '0',
author_id int(4) NOT NULL default '0'
);
INSERT INTO book_author (book_id, author_id)
VALUES (1,1), (1,2), (2,3), (4,1), (3,1), (5,2);
You want to list authors' books while suppressing repeating authors' names. A simple solution is to use MySQL's extremely useful GROUP_CONCAT() function to group books by author:
SELECT
a.name AS Author,
GROUP_CONCAT(b.name ORDER BY b.name) AS Books
FROM book_author AS ba
JOIN book AS b ON ba.book_id=b.id
JOIN author AS a ON ba.author_id=a.id
GROUP BY a.name;
For a neater-looking result:
1. Retrieve authors and their books.
2. Order them
3. Use a variable to remember and suppress repeating author names:
SET @last='';
SELECT
IF(r.author=@last,'',@last:=r.author) AS Author,
r.book AS Book
FROM (
SELECT DISTINCT a.name AS author,b.name AS book
FROM book_author AS ba
JOIN book AS b ON ba.book_id=b.id
JOIN author AS a ON ba.author_id=a.id
ORDER BY a.name,b.name
) AS r;
+---------------+---------------------------+
| author | book |
+---------------+---------------------------+
| Brad Phillips | Design Patterns |
| | MySQL in a bucket |
| | PHP Professional |
| Don Charles | Java Script Programming |
| | MySQL in a bucket |
| Kur Silver | Databases for Delinquents |
+---------------+---------------------------+
Pagination
Suppose you have a phone book of names, addresses, etc. You are displaying 20 rows per page, you're on page 100, and you want to display page 99. How do you do this knowing only what page you are on?
Assuming...
- 1-based page numbers
- you are on page P
- each page shows N rows
then the general formula for translating a 1-based page number into a first
LIMIT argument is ...
MAX(0,P-1) * N
which for the 99th 20-row page evaluates to 1960, and the second argument to
LIMIT is just N, so to see page 99, write...
SELECT ... LIMIT (1960, N);
The trouble with this is scaling. MySQL doesn't optimise
LIMIT well at all, so the bigger the table, the longer
LIMIT takes.
What's the alternative? Build pagination into the
WHERE clause, and ensure sure there is a covering index for the paginating column. On a table of 100,000 indexed random integers,
SELECT ... WHERE ... for the last 20 integers in the table is twice as fast as the comparable
LIMIT query. With a million integers, the ratio is over 500!
Pivot table basics: rows vs. columns
From table tbl( class, member ), you want to tabulate all classes and their members. In SQL terms, you aggregate members over classes. In MySQL:
SELECT class,GROUP_CONCAT(member)
FROM tbl
GROUP BY class;
With that simple query, you're halfway toward cross-tabulation, halfway to implementing a simple CUBE, and halfway to basic entity-attribute-value (EAV) logic. This is easier to see if we have two columns, rather than just one, to tabulate against the grouping column:
DROP TABLE IF EXISTS tbl;
CREATE TABLE tbl( id INT, colID INT, value CHAR(20) );
INSERT INTO tbl VALUES
(1,1,'Sampo'),(1,2,'Kallinen'),(1,3,'Office Manager'),
(2,1,'Jakko'),(2,2,'Salovaara'),(2,3,'Vice President');
To tabulate all colID and value values against all id values—that is, to write a reporting CUBE for the table—write a GROUP_CONCAT() instruction for each colID found in the table, then GROUP BY id:
SELECT
id,
GROUP_CONCAT(if(colID = 1, value, NULL)) AS 'First Name',
GROUP_CONCAT(if(colID = 2, value, NULL)) AS 'Last Name',
GROUP_CONCAT(if(colID = 3, value, NULL)) AS 'Job Title'
FROM tbl
GROUP BY id;
+------+------------+-----------+----------------+
| id | First Name | Last Name | Title |
+------+------------+-----------+----------------+
| 1 | Sampo | Kallinen | Office Manager |
| 2 | Jakko | Salovaara | Vice President |
+------+------------+-----------+----------------+
Since Paul Spinks first taught a spreadsheet how to do that cross-tabulation trick in 1979, this has also been known as a pivot table: we pivot colID and value against ID.
(Of course for a proper EAV representation, we'd add an attributes table:
DROP TABLE IF EXISTS attrs;
CREATE TABLE attrs(colID INT,attr CHAR(12));
INSERT INTO attrs VALUES (1,'First Name'),(2,'Last Name'),(3,'Title');
and write a stored procedure to PREPARE the above query from table-based attribute names.)
More often, crosstab queries calculate. Here is a simple sales table:
DROP TABLE IF EXISTS sales;
CREATE TABLE Sales (empID INT, yr SMALLINT, sales DECIMAL(10,2));
INSERT sales VALUES
(1, 2005, 12000),(1, 2006, 18000),(1, 2007, 25000),
(2, 2005, 15000),(2, 2006, 6000),(3, 2006, 20000),(3, 2007, 24000);
In Microsoft SQL Server, CUBE/PIVOT syntax for horizontal and vertical totals is:
SELECT EmpId, [2005], [2006], [2007], [ALL]
FROM (
SELECT
CASE WHEN GROUPING(EmpId)=0 THEN CAST(EmpId AS CHAR(7)) ELSE 'ALL' END AS EmpId,
CASE WHEN GROUPING(Yr)=0 THEN CAST(Yr AS CHAR(7)) ELSE 'ALL' END AS Yr,
SUM(Sales) AS Sales
FROM Sales
GROUP BY EmpId, Yr WITH CUBE
) AS s
PIVOT( SUM(Sales) FOR Yr IN ([2005], [2006], [2007], [ALL]) ) AS p
In MySQL, we write one aggregating expression per reporting column in the inner query, and build the horizontal sums in the outer query:
SELECT
IFNULL(empId,'Totals') AS EmpId,
sums.2005, sums.2006, sums.2007,
sums.2005 + sums.2006 + sums.2007 AS Sums
FROM (
SELECT
EmpID,
SUM(IF(Yr=2005,sales,0)) As '2005',
SUM(IF(Yr=2006,sales,0)) As '2006',
SUM(IF(Yr=2007,sales,0)) As '2007'
FROM Sales
GROUP BY EmpID WITH ROLLUP
) AS sums;
+--------+----------+----------+----------+-----------+
| EmpId | 2005 | 2006 | 2007 | Sums |
+--------+----------+----------+----------+-----------+
| 1 | 12000.00 | 18000.00 | 25000.00 | 55000.00 |
| 2 | 15000.00 | 6000.00 | 0.00 | 21000.00 |
| 3 | 0.00 | 20000.00 | 24000.00 | 44000.00 |
| Totals | 27000.00 | 44000.00 | 49000.00 | 120000.00 |
+--------+----------+----------+----------+-----------+
With many columns, pivot tables get complicated, time-consuming and error-prone. Automation is needed. Oracle and Microsoft SQL Server have CUBE syntax to simplify the job, especially for big pivot tables. MySQL doesn't. See "Automate pivot table queries" for how to roll your own cube with MySQL.
Last updated 20 Nov 2009
Automate pivot table queries
You have a
sales table listing product, salesperson and amount:
DROP TABLE IF EXISTS sales;
CREATE TABLE sales (
id int(11) default NULL,
product char(5) default NULL,
salesperson char(5) default NULL,
amount decimal(10,2) default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO sales VALUES
(1,'radio','bob','100.00'),
(2,'radio','sam','100.00'),
(3,'radio','sam','100.00'),
(4,'tv','bob','200.00'),
(5,'tv','sam','300.00'),
(6,'radio','bob','100.00');
SELECT * FROM sales;
+------+---------+-------------+--------+
| id | product | salesperson | amount |
+------+---------+-------------+--------+
| 1 | radio | bob | 100.00 |
| 2 | radio | sam | 100.00 |
| 3 | radio | sam | 100.00 |
| 4 | tv | bob | 200.00 |
| 5 | tv | sam | 300.00 |
| 6 | radio | bob | 100.00 |
+------+---------+-------------+--------+
If you are asked to tabulate sales amount against salesperson and product, you write a pivot table query:
SELECT
product,
SUM( CASE salesperson WHEN 'bob' THEN amount ELSE 0 END ) AS 'Bob',
SUM( CASE salesperson WHEN 'sam' THEN amount ELSE 0 END ) AS 'Sam',
SUM( amount ) AS Total
FROM sales
GROUP BY product WITH ROLLUP;
+---------+--------+--------+--------+
| product | Bob | Sam | Total |
+---------+--------+--------+--------+
| radio | 200.00 | 200.00 | 400.00 |
| tv | 200.00 | 300.00 | 500.00 |
| NULL | 400.00 | 500.00 | 900.00 |
+---------+--------+--------+--------+
The query generates one product per row and one column per salesperson. The pivoting
CASE expressions assign values of
sales.amount to the matching salesperson's column. For two products and two salespersons, it's a snap once you've done it a few times. When there are dozens of products and salespersons, though, writing the query becomes tiresome and error-prone.
Some years ago Giuseppe Maxia
published a little query that automates writing the pivot expressions. His idea was to embed the syntax for lines like the
SUM( CASE ...) lines above in a query for the
DISTINCT values. At the time Giuseppe was writing, MySQL did not support stored procedures. Now that it does, we can further generalise Giuseppe's idea by parameterising it in a stored procedure.
Admittedly, it's a little daunting. To write a query with variable names rather than the usual literal table and column names, we have to write
PREPARE statements. What we propose to do here is to write SQL that writes
PREPARE statements.
Code which writes code which writes code. Not a job for the back of a napkin.
It's easy enough to write the sproc shell. We keep generic queries in a
sys database, so the routine needs parameters specifying database, table, pivot column and (in some cases) the aggregating column. Then what? What worked for us was to proceed from back to front:
- Write the pivot expressions for a specific case.
- Write the
PREPARE statement that generates those expressions.
- Parameterise the result of #2.
- Put the result of #3 in an sproc.
Further complicating matters, we soon found that different summary aggregations, for example
COUNT and
SUM, require different sprocs. Here is the routine for generating
COUNT pivot expressions:
USE sys;
DROP PROCEDURE IF EXISTS writecountpivot;
DELIMITER |
CREATE PROCEDURE writecountpivot( db CHAR(64), tbl CHAR(64), col CHAR(64) )
BEGIN
DECLARE datadelim CHAR(1) DEFAULT '"';
DECLARE singlequote CHAR(1) DEFAULT CHAR(39);
DECLARE comma CHAR(1) DEFAULT ',';
SET @sqlmode = (SELECT @@sql_mode);
SET @@sql_mode='';
SET @sql = CONCAT( 'SELECT DISTINCT CONCAT(', singlequote,
',SUM(IF(', col, ' = ', datadelim, singlequote, comma,
col, comma, singlequote, datadelim, comma, '1,0)) AS `',
singlequote, comma, col, comma, singlequote, '`', singlequote,
') AS countpivotarg FROM ', db, '.', tbl,
' WHERE ', col, ' IS NOT NULL' );
-- UNCOMMENT TO SEE THE MIDLEVEL CODE:
-- SELECT @sql;
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
SET @@sql_mode=@sqlmode;
END;
|
DELIMITER ;
CALL sys.writecountpivot('test','sales','salesperson');
This generates the SQL ...
SELECT DISTINCT
CONCAT(',SUM(IF(salesperson = "',salesperson,'",1,0)) AS `',salesperson,'`')
AS countpivotarg
FROM test.sales
WHERE salesperson IS NOT NULL |
and returns...
+--------------------------------------------+
| countpivotarg |
+--------------------------------------------+
| ,SUM(IF(salesperson = "bob",1,0)) AS `bob` |
| ,SUM(IF(salesperson = "sam",1,0)) AS `sam` |
+--------------------------------------------+
which we plug into ...
SELECT
product
,SUM(IF(salesperson = "bob",1,0)) AS `bob`
,SUM(IF(salesperson = "sam",1,0)) AS `sam`
,COUNT(*) AS Total
FROM test.sales
GROUP BY product WITH ROLLUP;
+---------+------+------+-------+
| product | bob | sam | Total |
+---------+------+------+-------+
| radio | 2 | 2 | 4 |
| tv | 1 | 1 | 2 |
| NULL | 3 | 3 | 6 |
+---------+------+------+-------+
Not overwhelming for two columns, very convenient if there are 20. (Yes, it could also be written with
COUNT( ... 1, NULL)).
One point to notice is that the two levels of code generation create quotemark nesting problems. To make the double quotemark '"' available for data value delimiting, we turn off
ANSI_QUOTES during code generation, and put it back afterwards.
SUM pivot queries need different syntax:
USE sys;
DROP PROCEDURE IF EXISTS writesumpivot;
DELIMITER |
CREATE PROCEDURE writesumpivot( db CHAR(64), tbl CHAR(64), pivotcol CHAR(64), sumcol CHAR(64) )
BEGIN
DECLARE datadelim CHAR(1) DEFAULT '"';
DECLARE comma CHAR(1) DEFAULT ',';
DECLARE singlequote CHAR(1) DEFAULT CHAR(39);
SET @sqlmode = (SELECT @@sql_mode);
SET @@sql_mode='';
SET @sql = CONCAT( 'SELECT DISTINCT CONCAT(', singlequote,
',SUM(IF(', pivotcol, ' = ', datadelim, singlequote, comma,
pivotcol, comma, singlequote, datadelim, comma, sumcol, ',0)) AS `',
singlequote, comma, pivotcol, comma, singlequote, '`', singlequote,
') AS sumpivotarg FROM ', db, '.', tbl,
' WHERE ', pivotcol, ' IS NOT NULL' );
-- UNCOMMENT TO SEE THE MIDLEVEL SQL:
-- SELECT @sql;
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
SET @@sql_mode=@sqlmode;
END;
|
DELIMITER ;
CALL writesumpivot('test','sales','salesperson','amount');
+-------------------------------------------------+
| sumpivotarg |
+-------------------------------------------------+
| ,SUM(IF(salesperson = "bob",amount,0)) AS `bob` |
| ,SUM(IF(salesperson = "sam",amount,0)) AS `sam` |
+-------------------------------------------------+
which forms the guts of our report query:
SELECT
product
,SUM(IF(salesperson = "bob",amount,0)) AS `bob`
,SUM(IF(salesperson = "sam",amount,0)) AS `sam`
,SUM(amount) AS Total
FROM test.sales
GROUP BY product;
+---------+--------+--------+--------+
| product | bob | sam | Total |
+---------+--------+--------+--------+
| radio | 200.00 | 200.00 | 400.00 |
| tv | 200.00 | 300.00 | 500.00 |
+---------+--------+--------+--------+
There are higher levels of generality beckoning---say, a routine that generates a complete pivot table query, not just the pivot expressions.
Group column statistics in rows
A
pivot (or
crosstab, or
contingency) table aggregates sets of column values into rows of statistics, and
pivots target value statistics on partitioning criteria defined by any available data.
Spreadsheet applications have intuitive point-and-click interfaces for generating pivot tables. RDBMSs generally do not. The task looks difficult in SQL, though, only until you have coded a few.
If you ported the Microsoft sample database
Northwind to your MySQL database (as described in chapter 11 of
Get It Done with MySQL), you can execute this example step by step. Even if you haven't ported Northwind, the example is easy to follow.
Amongst the tables in the Northwind database are:
employees(employeeID, lastname, firstname, ...)
orders(orderID, customerId, employeeID, orderdate, ...)
There are nine employees, and 803 orders dated from 1996 through 1998. Each order points to an employeeID. Suppose we wish to report counts of orders taken by employees
pivoted on year--how would we proceed?
We do the pivot table 3-step:
1.
Write the basic aggregating query, a
GROUP BY query to aggregate the data on desired variables (in this case, employee and year).
2.
Write the pivoting query as an outer query that creates a column for each pivot value (year) from #1 written as an inner subquery. (Sometimes it is more efficient to write the results of #1 to a temp table and write #2 to refer to the temp table.)
3.
Fix a ROLLUP display glitch by encapsulating #2 in a new outer query that labels the
ROLLUP row meaningfully.
Here are the three steps in more detail:
1. Group the joined counts by the two criteria, employee and order year, yielding one result row per employee per year:
SELECT
CONCAT(firstname,' ',lastname) AS 'Employee',
YEAR(OrderDate) AS col,
COUNT(*) AS Data
FROM Employees e
JOIN Orders o ON e.EmployeeID = o.EmployeeID
GROUP BY e.employeeID, YEAR(o.OrderDate);
+------------------+------+------+
| Employee | col | Data |
+------------------+------+------+
| Nancy Davolio | 1996 | 26 |
| Nancy Davolio | 1997 | 55 |
| Nancy Davolio | 1998 | 42 |
| Andrew Fuller | 1996 | 16 |
| Andrew Fuller | 1997 | 41 |
| Andrew Fuller | 1998 | 39 |
| Janet Leverling | 1996 | 18 |
| Janet Leverling | 1997 | 71 |
| Janet Leverling | 1998 | 38 |
| Margaret Peacock | 1996 | 31 |
| Margaret Peacock | 1997 | 81 |
| Margaret Peacock | 1998 | 44 |
| Steven Buchanan | 1996 | 11 |
| Steven Buchanan | 1997 | 18 |
| Steven Buchanan | 1998 | 13 |
| Michael Suyama | 1996 | 15 |
| Michael Suyama | 1997 | 33 |
| Michael Suyama | 1998 | 19 |
| Robert King | 1996 | 11 |
| Robert King | 1997 | 36 |
| Robert King | 1998 | 25 |
| Laura Callahan | 1996 | 19 |
| Laura Callahan | 1997 | 54 |
| Laura Callahan | 1998 | 31 |
| Anne Dodsworth | 1996 | 5 |
| Anne Dodsworth | 1997 | 19 |
| Anne Dodsworth | 1998 | 19 |
+------------------+------+------+
Nine employees for three years yield 27 aggregated rows.
2. We want one summary row per employee, and one count column for each year when an employee took an order. We
pivot the rows of the above resultset on year by querying the above resultset, defining a colunn for every
year found, for example:
SUM( CASE col WHEN '1996' THEN data ELSE 0 END ) AS '1996',
grouping the result by
row WITH ROLLUP to provide a row of column sums at the bottom. This gives the following query:
SELECT
Employee,
SUM( CASE col WHEN '1996' THEN data ELSE 0 END ) AS '1996',
SUM( CASE col WHEN '1997' THEN data ELSE 0 END ) AS '1997',
SUM( CASE col WHEN '1998' THEN data ELSE 0 END ) AS '1998',
SUM( data ) AS Total -- sums across years by employee
FROM (
SELECT -- the query from step #1
CONCAT(firstname,' ',lastname) AS 'Employee',
YEAR(OrderDate) AS 'col',
COUNT(*) AS Data
FROM Employees e
JOIN Orders o ON e.EmployeeID = o.EmployeeID
GROUP BY e.employeeID, YEAR(o.OrderDate)
) AS stats
GROUP BY employee WITH ROLLUP;
+------------------+------+------+------+-------+
| Employee | 1996 | 1997 | 1998 | Total |
+------------------+------+------+------+-------+
| Andrew Fuller | 16 | 41 | 39 | 96 |
| Anne Dodsworth | 5 | 19 | 19 | 43 |
| Janet Leverling | 18 | 71 | 38 | 127 |
| Laura Callahan | 19 | 54 | 31 | 104 |
| Margaret Peacock | 31 | 81 | 44 | 156 |
| Michael Suyama | 15 | 33 | 19 | 67 |
| Nancy Davolio | 26 | 55 | 42 | 123 |
| Robert King | 11 | 36 | 25 | 72 |
| Steven Buchanan | 11 | 18 | 13 | 42 |
| NULL | 152 | 408 | 270 | 830 |
+------------------+------+------+------+-------+
3. The result of #2 is correct except that sums ought not to be reported as NULL! We fix that bit of weirdness by writing query #2 as a derived table, and having the new outer query alias the yearly sums row:
SELECT
IFNULL( employee, 'SUMS') AS Employee, 1996, 1997, 1998, Total
FROM (
SELECT
Employee,
SUM( CASE col WHEN '1996' THEN data ELSE 0 END ) AS '1996',
SUM( CASE col WHEN '1997' THEN data ELSE 0 END ) AS '1997',
SUM( CASE col WHEN '1998' THEN data ELSE 0 END ) AS '1998',
SUM( data ) AS Total
FROM (
SELECT
CONCAT(firstname,' ',lastname) AS 'Employee',
YEAR(OrderDate) AS 'col',
COUNT(*) AS Data
FROM Employees e
JOIN Orders o ON e.EmployeeID = o.EmployeeID
GROUP BY e.employeeID, YEAR(o.OrderDate)
) AS stats
GROUP BY employee WITH ROLLUP
) AS stats2;
+------------------+------+------+------+-------+
| Employee | 1996 | 1997 | 1998 | Total |
+------------------+------+------+------+-------+
| Andrew Fuller | 1996 | 1997 | 1998 | 96 |
| Anne Dodsworth | 1996 | 1997 | 1998 | 43 |
| Janet Leverling | 1996 | 1997 | 1998 | 127 |
| Laura Callahan | 1996 | 1997 | 1998 | 104 |
| Margaret Peacock | 1996 | 1997 | 1998 | 156 |
| Michael Suyama | 1996 | 1997 | 1998 | 67 |
| Nancy Davolio | 1996 | 1997 | 1998 | 123 |
| Robert King | 1996 | 1997 | 1998 | 72 |
| Steven Buchanan | 1996 | 1997 | 1998 | 42 |
| Sums | 1996 | 1997 | 1998 | 830 |
+------------------+------+------+------+-------+
With multiple statistics and pivot layers, a pivot table query can get complex, but following this 3-step will keep things clear.
Pivot table using math tricks
http://en.wikibooks.org/wiki/Programming:MySQL/Pivot_table
Pivot table with CONCAT
Here is a MySQL pivot table query for room bookings by weekday:
SELECT slot
, max(if(day=1, concat(subject,' ',room), '')) as day1
, max(if(day=2, concat(subject,' ',room), '')) as day2
, max(if(day=3, concat(subject,' ',room), '')) as day3
, max(if(day=4, concat(subject,' ',room), '')) as day4
, max(if(day=5, concat(subject,' ',room), '')) as day5
from schedule
group by slot
MAX(...) decides between an entry and a blank (the entry will win if one exists) while the group by lines everything up on the same row. Friendly caution: If more than one entry exists for the same day and time, you will only see the one that is alphabetically "greater".
To see how many classes are scheduled by day for each slot (to check for conflicts) try:
SELECT slot
, sum(if(day=1,1,0)) as day1
, sum(if(day=2,1,0)) as day2
, sum(if(day=3,1,0)) as day3
, sum(if(day=4,1,0)) as day4
, sum(if(day=5,1,0)) as day5
from schedule
group by slot
There is a pattern:
- Columns you want as "row headers" are listed both in the SELECT _and_ in the GROUP BY clauses
- Values you want as columns are selectively chosen by IF() functions to return something or nothing so that one of the aggregate functions (MIN, MAX, SUM,AVG, etc) can condense those rows and columns into single values
What to count in any column can be as complex as you like. If you don't like the IF() function you can use CASE statements or anything else to help you provide the correct set of values for the aggregate function you are using on that column. You can mix and match aggregate functions to be what you need. Let's say you join the appropriate tables together to form a data set that looks like:
day, slot, subject, student, grade
and you wanted to build a chart showing the statistics of grades vs. days and subject. You could use this type of query to work that out:
SELECT day, subject
, AVG(grade) as average
, MIN(grade) as lowest
, MAX(grade) as highest
from <necessary tables>
group by day, subject
Now while one needn't choose values for columns, to "pivot" that table by days (each column representing statistics for just one day) change the query to:
SELECT subject
, AVG(IF(day=1, grade,null)) as D1_average
, MIN(IF(day=1, grade, null)) as D1_lowest
, MAX(IF(day=1,grade,null)) as D1_highest
, AVG(IF(day=2, grade,null)) as D2_average
, MIN(IF(day=2, grade, null)) as D2_lowest
, MAX(IF(day=2,grade,null)) as D2_highest
, .... (repeat for rest of the days)
FROM <necessary tables>
GROUP BY day, subject
The IF ... NULL test prevents AVG() from counting all other grades for the same subject from different days. The same trick works for MIN and MAX functions.
Pivot table without GROUP_CONCAT
Data designs often require flexibility in numbers and names of data points per instance row: instead of saving all the data points belonging to a key value in a single row, you save each data point as a name-value pair in its own row.
Thus given table user_class(user_id INT, class_id CHAR(20), class_value CHAR(20)) with these rows:
user_id class_id class_value
1 firstname Rogier
1 lastname Marat
2 firstname Jean
2 lastname Smith
and you wish a resultset that links first names to last names for each ID...
user_id firstname lastname
1 Rogier Marat
2 Jean Smith
the following query accomplishes the required pivot via an INNER SELF-JOIN:
SELECT
u1.user_ID,
class_value AS firstname,
u2.lastname
FROM user_class AS u1
INNER JOIN (
SELECT
user_ID,
class_value AS lastname
FROM user_class
WHERE class_id='lastname'
) AS u2
ON u1.user_ID=u2.user_ID AND u1.class_id='firstname'
All possible recipes with given ingredients
We have tables for recipes (r), ingredients required for recipes (ri), and ingredients now available in the pantry (p). In table p there may be many rows for a given recipe, each specifying one ingredient.
drop table if exists r,p,ri;
create table r(id int);
insert into r values(1),(2),(3);
create table p(id int);
insert into p values(1),(2),(3);
create table ri(rid int,pid int);
insert into ri values (1,1),(1,2),(2,1),(2,4),(3,5),(3,6),(3,7);
select id as recipes from r;
+---------+
| recipes |
+---------+
| 1 |
| 2 |
| 3 |
+---------+
select id as 'available ingredients' from p;
+-----------------------+
| available ingredients |
+-----------------------+
| 1 |
| 2 |
| 3 |
+-----------------------+
select rid as recipe, pid as ingredient from ri;
+--------+------------+
| recipe | ingredient |
+--------+------------+
| 1 | 1 |
| 1 | 2 |
| 2 | 1 |
| 2 | 4 |
| 3 | 5 |
| 3 | 6 |
| 3 | 7 |
+--------+------------+
Given our ingredients, what recipes can we make? Inspection shows the answer is recipe #1.
SQL has no universal quantifier, so how do we proceed? 'All A is B' is logically equivalent to the double negative 'there is no A that is not B', so we can reformulate the requirement ...
list the recipes for which we have all ingredients
into terms SQL can handle ...
list the recipes for which there is no ingredient we do not have
A double negative, so a double query. One inner query, one outer. Tackle the inner one first: find the recipes for which we are missing an ingredient.
That's a straight exclusion join, i.e., a left join on ingredient from 'required' to 'available', plus a where clause that restricts the resultset to nulls on the right ('available') side of the join:
SELECT DISTINCT rid AS 'Recipes for which the pantry is missing some ingredients'
FROM ri
LEFT JOIN p ON ri.pid=p.id
WHERE p.id IS NULL;
+----------------------------------------------------------+
| Recipes for which the pantry is missing some ingredients |
+----------------------------------------------------------+
| 2 |
| 3 |
+----------------------------------------------------------+
Our outer query has to find the recipes which are not in this list. That's another exclusion join, this time from recipes to the above derived table:
SELECT r.id
FROM r
LEFT JOIN (
SELECT DISTINCT rid
FROM ri
LEFT JOIN p ON ri.pid=p.id
WHERE p.id IS NULL
) AS rno ON r.id = rno.rid
WHERE rno.rid IS NULL;
+------+
| id |
+------+
| 1 |
+------+
It's an example of relational division, one of Codd's eight basic relational operations. Dividing a divisor table into a dividend table yields a quotient or results table:
dividend ÷ divisor = quotient
As in arithmetic, multiplication reverses it:
divisor * quotient = dividend
+-----------+
+-----+ +------+ | table AxB |
| A | | B | +-----+-----+
+-----+ +------+ |key_a|key_b|
|key_a| |key_b | +-----+-----+
+-----+ +------+ | 2 | 1 |
| 2 | | 1 | | 2 | 7 |
| 4 | | 7 | | 2 | 3 |
+-----+ | 3 | | 4 | 1 |
+------+ | 4 | 7 |
| 4 | 3 |
+-----+-----+
When we multiply (CROSS JOIN) tables A and B to yield AxB, AxB gets a row combining every row of A with every row of B, and all the columns from A and B. When we reverse that operation, dividing AxB by B, we get back A by listing distinct B values associated with A values in AxB.
All X for which all Y are Z (relational division)
You have an election database with tables listing political
parties, election
districts, and
candidates running for parties in those districts. You want to know which parties have candidates running in all districts. Under
Aggregates we show a
GROUP BY solution (
here).
If there are reasons not to aggregate,
relational division can solve the problem. The basic idea in relational division is that, aside from aggregation, SQL has no direct way to express "all Xs for which all Y are Z", but does have a
NOT EXISTS operator, so we can express "all Xs for which all Y are Z" in SQL as a double negative: "all Xs for which no Y is not Z". Once you think of formulating the question this way, the query almost writes itself:
SELECT DISTINCT party FROM parties
WHERE NOT EXISTS (
SELECT * FROM districts
WHERE NOT EXISTS (
SELECT * FROM candidates
WHERE candidates.party=parties.party AND candidates.district=districts.district
)
);
Why is it called relational division? See the
All possible recipes with given ingredients entry. Here the dividend is
candidates, the divisor is
districts and the quotient is a party count.
Most
NOT EXISTS() queries can be translated into exclusion joins, which are often much faster. An exclusion join from A to B excludes A rows for which the
LEFT JOIN condition finds
NULLs in B. The query we are translating has two
NOT EXISTS clauses, so we need two exclusion joins:
SELECT p.party
FROM parties p
LEFT JOIN (
SELECT a.party
FROM (
SELECT DISTINCT party,district
FROM parties CROSS JOIN districts
) a
LEFT JOIN candidates c ON a.party=c.party AND a.district=c.district
WHERE c.party IS NULL
) b ON p.party=b.party
WHERE b.party IS NULL;
Like numeric division, relational division has a gotcha:
divide by zero. If the divisor table has zero rows, the quotient counts
all distinct dividend instances. If that is not what you want, use
aggregation.
Most "all Xs for which all Y are Z" queries can be written in any of these three ways. Try each one to see which performs best for your problem.
Who makes all the parts for a given assembly?
One way to arrive at the answer is by asking: What are the assembly-supplier pairs such that no part of the assembly is not made by the supplier? That's relational division again, formulated for two tables by Stephen Todd. Given assemblyparts(assembly,part) and partsuppliers(part,supplier) tables, here is a query that Joe Celko credits to Pierre Mullin.
SELECT DISTINCT
AP1.assembly,
SP1.supplier
FROM AssemblyParts AS AP1, PartSuppliers AS SP1
WHERE NOT EXISTS (
SELECT *
FROM AssemblyParts AS AP2
WHERE AP2.assembly = AP1.assembly
AND NOT EXISTS (
SELECT SP2.part
FROM PartSuppliers AS SP2
WHERE SP2.part = AP2.part AND SP2.supplier = SP1.supplier
)
);
Find blocks of unused numbers
In a table of sequential IDs with no missing values, some are used and some are not. Find the blocks of unused IDs, if any:
DROP TABLE IF EXISTS tbl;
CREATE TABLE tbl(id INT,used BOOL);
INSERT INTO tbl VALUES (1,1),(2,1),(3,0),(4,1),(5,0),(6,1),(7,1),(8,1),
(9,0),(10,0),(11,1),(12,1),(13,0),(14,0),(15,0);
SELECT * FROM tbl;
+------+------+
| id | used |
+------+------+
| 1 | 1 |
| 2 | 1 |
| 3 | 0 |
| 4 | 1 |
| 5 | 0 |
| 6 | 1 |
| 7 | 1 |
| 8 | 1 |
| 9 | 0 |
| 10 | 0 |
| 11 | 1 |
| 12 | 1 |
| 13 | 0 |
| 14 | 0 |
| 15 | 0 |
+------+------+
The first ID in any unused sequence has used=0 and either no immediate predecessor, or an immediate predecessor where used=1. The last ID of any unused sequence either has no successor or the successor has used=1. So:
1. Find the first first ID of every unused sequence by left joining each row with used=0 to the immediate predecessor row, conditioning the result on the predecessor row not existing or having used=1.
2. As a basis for finding the last ID of every unused sequence that is followed by a row with used=1, left join first unused rows to rows with larger IDs and used=1.
3. As a basis for finding the last ID of an unused sequence which is also the largest ID in the table, left join first unused rows to rows with larger IDs and used=0.
4. For each first unused ID, the last unused ID in its sequence is one less than the smallest used ID greater than the first ID if it exists, otherwise it is the maximum unused ID greater than the first ID.
SELECT firstUnused, IF(mincid IS NULL, IFNULL(did,firstUnused),mincid-1) AS lastUnused
FROM (
SELECT first.id AS firstUnused, MIN(c.id) AS mincid, MAX(d.id) AS did
FROM (
SELECT a.id
FROM tbl a
LEFT JOIN tbl b ON a.id=b.id + 1
WHERE a.used=0 AND (b.id IS NULL OR b.used=1)
) AS first
LEFT JOIN tbl c ON first.id<c.id AND c.used=1
LEFT JOIN tbl d ON first.id<d.id AND d.used=0
GROUP BY firstUnused
) AS e;
+-------------+------------+
| firstUnused | lastUnused |
+-------------+------------+
| 3 | 3 |
| 5 | 5 |
| 9 | 10 |
| 13 | 15 |
+-------------+------------+
Thanks to Don Armstrong for finding a case where our previous algorithm failed.
Last updated 25 Sep 2009
Find missing numbers in a sequence
You have a table tbl(id int) with values (1,2,4,18,19,20,21), and you wish to find the first missing number in its sequence of id values:
SELECT t1.id+1 AS Missing
FROM tbl AS t1
LEFT JOIN tbl AS t2 ON t1.id+1 = t2.id
WHERE t2.id IS NULL
ORDER BY id LIMIT 1;
+---------+
| Missing |
+---------+
| 3 |
+---------+
For all the gaps, including gaps of more than 1 value, you need something a little more baroque...
SELECT
a.id+1 AS 'Missing From',
MIN(b.id) - 1 AS 'To'
FROM tbl AS a, tbl AS b
WHERE a.id < b.id
GROUP BY a.id
HAVING `Missing From` < MIN(b.id);
+--------------+------+
| Missing From | To |
+--------------+------+
| 3 | 3 |
| 5 | 17 |
+--------------+------+
We often need such lists, so the query is a natural for a stored procedure that finds missing sequence values in any table:
DROP PROCEDURE IF EXISTS MissingInSeq;
DELIMITER |
CREATE PROCEDURE MissingInSeq( db VARCHAR(64), tbl VARCHAR(64), col VARCHAR(64) )
BEGIN
SET @sql = CONCAT( "SELECT a.", col,
"+1 AS 'Missing From',",
"MIN(b.",
col,
") - 1 AS 'To' FROM ",
db,
".",
tbl,
" AS a,",
db,
".",
tbl,
" AS b WHERE a.",
col,
" < b.",
col,
" GROUP BY a.",
col,
" HAVING a.",
col,
" < MIN(b.",
col,
") - 1"
);
-- SELECT @sql;
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
END;
|
DELIMITER ;
Find previous and next values in a sequence
Given a table
t(ID int, ...), how would we display each ID and its next highest value? A simple method uses aggregation on a non-equi-self-join to find the smallest value of all greater values for each value:
SELECT id, MIN(b.id) AS Next
FROM t AS a
JOIN t AS b ON b.id > a.id
GROUP BY a.id;
The same logic can retrieve next lowest values.
Suppose we wish to track daily changes in closing price:
drop table if exists t;
create table t (date date, closing_price decimal(8,2) );
insert into t values
('2009-01-02', 5),('2009-01-03', 5.5),('2009-01-04', 4),('2009-01-05', 6);
We can use the above non-equi-self-join to compute the ratios:
SELECT
seq.Next AS date, t.closing_price/seq.closing_price AS ChangeRatio
FROM t
JOIN (
SELECT a.date, a.closing_price, MIN(b.date) AS Next
FROM t AS a
JOIN t AS b ON b.date > a.date
GROUP BY a.date
) seq ON t.date=seq.next;
+------------+-------------+
| date | ChangeRatio |
+------------+-------------+
| 2009-01-03 | 1.100000 |
| 2009-01-04 | 0.727273 |
| 2009-01-05 | 1.500000 |
+------------+-------------+
Here is another algorithm, by Baron Schwartz (
xaprb.com), for retrieving the previous and next column values in a sequence, given a particular column value
thisvalue. The previous value is the maximum value less than
thisvalue, and the next value is the minimum value greater than
thisvalue:
SELECT
IF(col > thisvalue,'next','prev') AS Direction,
IF(col > thisvalue,MIN(col),MAX(col)) AS 'Prev/Next'
FROM tablename
WHERE col <> thisvalue
GROUP BY SIGN(col - thisvalue);
So, to find the previous and next order ids in the Northwind database table orders (nwib.orders), starting from order number 10800:
SELECT
IF(orderid > 10800,'next','prev') AS Direction,
IF(orderid > 10800,MIN(orderid),MAX(orderid)) AS 'Prev/Next'
FROM nwib.orders
WHERE orderid <> 10800
GROUP BY SIGN(orderid - 10800);
+-----------+-----------+
| Direction | Prev/Next |
+-----------+-----------+
| prev | 10799 |
| next | 10801 |
+-----------+-----------+
This is a natural for a stored procedure:
DROP PROCEDURE IF EXISTS PrevNext;
DELIMITER |
CREATE PROCEDURE PrevNext(
IN db CHAR(64), IN tbl CHAR(64), IN col CHAR(64), IN seq INT
)
BEGIN
IF db IS NULL OR db = '' THEN
SET db = SCHEMA();
END IF;
SET @sql = CONCAT( "SELECT ",
" IF(", col, " > ", seq,",'next','prev') AS Direction,",
" IF(", col, " > ", seq, ",MIN(", col, "),MAX(", col, ")) AS 'Prev/Next'",
" FROM ", db, ".", tbl,
" WHERE ", col, " <> ", seq,
" GROUP BY SIGN(", col, " - ", seq, ")" );
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
END;
|
DELIMITER ;
Or, it can be embedded in the
FROM clause of another query, for example ...
SELECT o2.OrderID,o2.Value,o.customerid
FROM orders o
JOIN (
SELECT 'This' AS 'OrderId', 10800 AS 'Value'
UNION
SELECT
IF( orderid > 10800, 'Next', 'Prev') AS Which,
IF( orderid > 10800, MIN(orderid), MAX(orderid )) AS 'Value'
FROM orders
WHERE orderid <> 10800
GROUP BY SIGN( orderid - 10800 )
) AS o2 ON o.orderid=o2.value
ORDER BY o.orderid;
+---------+-------+------------+
| OrderID | Value | customerid |
+---------+-------+------------+
| Prev | 10799 | KOENE |
| This | 10800 | SEVES |
| Next | 10801 | BOLID |
+---------+-------+------------+
Find row with next value of specified column
Sometimes we need next values of a column on a given row ordering. Oracle has a LEAD(...) OVER(...) construct which simplifies this query. MySQL does not. The logic is:
1. Form a resultset consisting of all relevant rows joined with all relevant rows that have greater values in the ordering columns. For example, if the table has these rows:
+------+
| 2 |
| 4 |
| 6 |
| 8 |
| 10 |
+------+
Then the resultset is
+------+------+
| 2 | 4 |
| 2 | 6 |
| 2 | 8 |
| 2 | 10 |
| 4 | 6 |
| 4 | 8 |
| 4 | 10 |
| 6 | 8 |
| 6 | 10 |
| 8 | 10 |
+------+------+
2. For each row in this resultset, find the row with the lowest ordering value amongst the higher values. For the example the result of this from the above resultset is
+-----+-----+
| 2 | 4 |
| 4 | 6 |
| 6 | 8 |
| 8 | 10 |
+-----+-----+
The fastest way to do step #2 is a self-exclusion join (see self-exclusion join examples elsewhere on this page).
Put it together:
drop table if exists t;
create table t(id int);
insert into t values(2),(4),(6),(8),(10);
select x.aid as id,x.bid as nextvalue
from (
select a.id as aid,b.id as bid
from t a
join t b on a.id<b.id
) x
left join (
select a.id as aid,b.id as bid
from t a
join t b on a.id<b.id
) y on x.aid=y.aid and x.bid>y.bid
where y.bid is null
order by x.aid,x.bid;
+------+-----------+
| id | nextvalue |
+------+-----------+
| 2 | 4 |
| 4 | 6 |
| 6 | 8 |
| 8 | 10 |
+------+-----------+
Modify the algorithm to suit for next lowest. Modify the algorithm to suit for next lowest. Here is an example where the sequencing column is a timestamp:
drop table if exists t;
create table t(d timestamp,i int);
insert into t values
('2009-12-1 00:00:00',1),('2009-12-3 00:00:00',3),('2009-12-5 00:00:00',5),('2009-12-8 00:00:00',8);
select x.*
from (
select a.d as thisdate, a.i as thisvalue, b.d as nextdate, b.i as nextvalue
from t a
join t b on a.d < b.d
) x
left join (
select a.d as thisdate, b.d as nextdate
from t a
join t b on a.d < b.d
) y on x.thisdate = y.thisdate and x.nextdate > y.nextdate
where y.nextdate is null
order by x.thisdate, x.nextdate;
+---------------------+-----------+---------------------+-----------+
| thisdate | thisvalue | nextdate | nextvalue |
+---------------------+-----------+---------------------+-----------+
| 2009-12-01 00:00:00 | 1 | 2009-12-03 00:00:00 | 3 |
| 2009-12-03 00:00:00 | 3 | 2009-12-05 00:00:00 | 5 |
| 2009-12-05 00:00:00 | 5 | 2009-12-08 00:00:00 | 8 |
+---------------------+-----------+---------------------+-----------+
Last updated 11 Dec 2009
Find sequence starts and ends
To find the first and last values of column value sequences in a table like this ...
drop table if exists t;
create table t(id int);
insert into t values(1),(2),(3),(4),(6),(7),(8);
an exclusion join on the previous sequential value finds the first value of each sequence, and the minimum next value from a left join and an exclusion join on the previous sequential value finds the end of each sequence:
SELECT
a.id AS Start,
MIN( c.id ) AS End
FROM tbl AS a
LEFT JOIN tbl AS b ON a.id = b.id + 1
LEFT JOIN tbl AS c ON a.id <= c.id
LEFT JOIN tbl AS d ON c.id = d.id - 1
WHERE b.id IS NULL
AND c.id IS NOT NULL
AND d.id IS NULL
GROUP BY a.id;
+-------+------+
| Start | End |
+-------+------+
| 1 | 4 |
| 6 | 8 |
+-------+------+
Thanks to Scott Noyes for noticing that a.id<c.id fails to pick up sequences of 1 followed by skips of 1, but a.id<=c.id does.
To see how that query works, look at the output of this version of the query with exclusion and aggregation clauses removed:
SELECT a.id AS aid,b.id AS bid, c.id AS c.id, d.di AS did
FROM tbl AS a
LEFT JOIN tbl AS b ON a.id = b.id + 1
LEFT JOIN tbl AS c ON a.id <= c.id
LEFT JOIN tbl AS d ON c.id = d.id - 1
ORDER BY a.id,b.id,c.id,d.id;
+------+------+------+------+
| aid | bid | cid | did |
+------+------+------+------+
| 1 | NULL | 1 | 2 |
| 1 | NULL | 2 | 3 |
| 1 | NULL | 3 | 4 |
| 1 | NULL | 4 | NULL | <-- end of sequence starting with 1
| 1 | NULL | 6 | 7 |
| 1 | NULL | 7 | 8 |
| 1 | NULL | 8 | NULL |
| 2 | 1 | 2 | 3 |
| 2 | 1 | 3 | 4 |
| 2 | 1 | 4 | NULL |
| 2 | 1 | 6 | 7 |
| 2 | 1 | 7 | 8 |
| 2 | 1 | 8 | NULL |
| 3 | 2 | 3 | 4 |
| 3 | 2 | 4 | NULL |
| 3 | 2 | 6 | 7 |
| 3 | 2 | 7 | 8 |
| 3 | 2 | 8 | NULL |
| 4 | 3 | 4 | NULL |
| 4 | 3 | 6 | 7 |
| 4 | 3 | 7 | 8 |
| 4 | 3 | 8 | NULL |
| 6 | NULL | 6 | 7 |
| 6 | NULL | 7 | 8 |
| 6 | NULL | 8 | NULL | <-- end of sequence starting with 6
| 7 | 6 | 7 | 8 |
| 7 | 6 | 8 | NULL |
| 8 | 7 | 8 | NULL |
+------+------+------+------+
A variant of the problem: when some IDs are used and some are not, find blocks of unused IDs:
DROP TABLE IF EXISTS tbl;
CREATE TABLE tbl(id INT,used BOOL);
INSERT INTO tbl VALUES(1,1),(2,0),(3,0),(4,1),(5,0),(6,0);
SELECT a.id AS Start, MIN( c.id ) AS End
FROM tbl AS a
LEFT JOIN tbl AS b ON a.id=b.id + 1 AND a.used=0 AND b.used=0
LEFT JOIN tbl AS c ON a.id<=c.id AND a.used=0 AND c.used=0
LEFT JOIN tbl AS d ON c.id=d.id-1 AND c.used=0 AND d.used=0
WHERE b.id IS NULL
AND c.id IS NOT NULL
AND d.id IS NULL
GROUP BY a.id;
+-------+------+
| Start | End |
+-------+------+
| 2 | 3 |
| 5 | 6 |
+-------+------+
Last updated 04 Feb 2010
Find specific sequences
You have a table which tracks hits on named web pages...
CREATE TABLE hits (
id INT NOT NULL DEFAULT 0,
page CHAR(128) DEFAULT '',
time TIMESTAMP NOT NULL DEFAULT 0,
PRIMARY KEY(id, time)
)
where id is unique to a session. Here is a bit of sample data:
INSERT INTO hits VALUES
(1, 'A', TIMESTAMPADD(SECOND,10,NOW())),
(1, 'B', TIMESTAMPADD(SECOND,20,NOW())),
(2, 'A', TIMESTAMPADD(SECOND,40,NOW())),
(1, 'A', TIMESTAMPADD(SECOND,50,NOW())),
(1, 'C', TIMESTAMPADD(SECOND,60,NOW())),
(3, 'A', TIMESTAMPADD(SECOND,110,NOW())),
(3, 'A', TIMESTAMPADD(SECOND,120,NOW())),
(3, 'C', TIMESTAMPADD(SECOND,130,NOW())),
(2, 'C', TIMESTAMPADD(SECOND,90,NOW())),
(2, 'A', TIMESTAMPADD(SECOND,100,NOW()));
You desire a count of the number of sessions where a user moved from one particular page directly to another, for example from 'A' to 'C'.
To find the next hit in a given session, scope on id, order by time, and limit the output to one row. Then simply count the rows meeting the page criteria:
SELECT
COUNT(DISTINCT h1.id) AS 'Moves from A to C'
FROM hits AS h1
WHERE
h1.page = 'A'
AND 'C' = (
SELECT h2.page
FROM hits AS h2
WHERE h2.id = h1.id
AND h2.time > h1.time
ORDER BY h2.time LIMIT 1
);
-------------------
| Moves from A to C |
-------------------
| 3 |
-------------------
Gaps in a time series
Advanced time series analysis generally requires custom software, but straightforward SQL queries can answer simple time series questions. You have a jobtimes table with columns ID, job, machine, start_time, and stop_time. You wish to know which machines have had gaps between activity periods.Here is a query that shows the start times following breaks in activity for a given machine.
SELECT
id,
machine AS thismachine,
start_time AS StartAfterGap
FROM jobtimes
WHERE id > 1 AND NOT EXISTS (
SELECT stop_time
FROM jobtimes
WHERE machine=thismachine
AND start_time < StartAfterGap
AND stop_time >= StartAfterGap
)
Make values of a column sequential
You have a table tbl with an integer primary key column keycol which is not a key in another table, and which you wish to make perfectly sequential starting with 1.
SET @i=0;
UPDATE tbl SET keycol=(@i:=@i+1);
Track stepwise project completion
A master table has one row for each project, and the number of sequential steps required to complete each project. A detail table has one row per project per completed step:
DROP TABLE IF EXISTS t1 ;
CREATE TABLE t1 (
id INT, projectname CHAR(2), projectsteps INT
);
INSERT INTO t1 VALUES
(1, 'xx', 3),
(2, 'yy', 3),
(3, 'zz', 5);
DROP TABLE IF EXISTS t2;
CREATE TABLE t2 (
id INT, projectID INT, xid INT
);
INSERT INTO t2 VALUES
(1, 1, 1),
(2, 1, 2),
(3, 2, 1),
(4, 1, 3),
(5, 3, 2),
(6, 1, 2),
(7, 2, 1),
(8, 2, 1);
The requirement is for a query which, for every project, reports 'OK' if there is at least one detail row for every project step, or otherwise reports the number of the last sequential completed step:
Here is one way to build such a query:
1. Join t1 to t2 on projectID.
2. Left Join t2 to itself on projectID and integer succession.
3. Add a WHERE condition which turns the left self-join into an exclusion join that finds the first missing sequential xid value.
4. To the SELECT list add this item:
IF( a.xid < p.projectstep,a.xid,'OK' ) AS StepState
so when the first sequential missing xid is not less than the number of project steps, display 'Ok', otherwise display the xid value before the first missing xid value.
5. Remove dupes with a GROUP BY clause.
SELECT
p.projectname,p.projectsteps,a.xid,
IF(a.xid < p.projectsteps, a.xid, 'OK') AS CompletionState
FROM t1 p
JOIN t2 a ON p.id = a.projectID
LEFT JOIN t2 AS b ON a.projectID = b.projectID AND a.xid+1 = b.xid
WHERE b.xid IS NULL
GROUP BY p.projectname;
+-------------+--------------+------+-----------------+
| projectname | projectsteps | xid | CompletionState |
+-------------+--------------+------+-----------------+
| xx | 3 | 3 | OK |
| yy | 3 | 1 | 1 |
| zz | 5 | 2 | 2 |
+-------------+--------------+------+-----------------+
Winning Streaks
Given a table of IDs and won-lost results, how do we find the longest winning streak?
drop table if exists results;
create table results(id int,result char(1));
insert into results values
(1,'w'),(2,'l'),(3,'l'),(4,'w'),(5,'w'),(6,'w'),(7,'l'),(8,'w'),(9,'w');
select * from results;
+------+--------+
| id | result |
+------+--------+
| 1 | w |
| 2 | l |
| 3 | l |
| 4 | w |
| 5 | w |
| 6 | w |
| 7 | l |
| 8 | w |
| 9 | w |
+------+--------+
We can find streaks of two with a left join on a.id=b.id+1. To count streak lengths, initialise a counter to 0 then increment it for every hit:
set @count=0;
select a.id, a.result, b.result, @count := IF(a.result = b.result, @count + 1, 1) as Streak
from results a
left join results b on a.id = b.id + 1
where a.result = 'w';
The longest winning streak is the longest such streak found:
set @count=0;
select MAX(@count:=IF(a.result = b.result, @count + 1, 1)) as LongestStreak
from results a
left join results b on a.id = b.id + 1
where a.result = 'w';
+---------------+
| LongestStreak |
+---------------+
| 3 |
+---------------+
That solution is from a response by Jon Roshko to a question by Ed Ball on the MySQL Newbie Forum. Scott Noyes points out that our query pattern for sequence starts and ends also works for winning streaks:
SELECT MIN( c.id ) - a.id + 1 as LongestStreak
FROM results AS a
LEFT JOIN results AS b ON a.id = b.id + 1 AND b.result = 'w'
LEFT JOIN results AS c ON a.id <= c.id AND c.result = 'w'
LEFT JOIN results AS d ON c.id = d.id - 1 AND d.result = 'w'
WHERE
a.result = 'w'
AND b.id IS NULL
AND c.id IS NOT NULL
AND d.id IS NULL
GROUP BY a.id
ORDER BY LongestStreak DESC LIMIT 1;
Great circle distance
Find the distance in kilometres between two points on the surface of the earth. This is just the sort of problem stored functions were made for. For a first order approximation, ignore deviations of the earth's surface from the perfectly spherical. Then the distance in radians is given by a number of trigonometric formulas. ACOS and COS behave reasonably:
COS(lat1-lat2)*(1+COS(lon1-lon2)) - COS(lat1+lat2)*(1-COS(lon1-lon2))
rads = ACOS( --------------------------------------------------------------------- )
2
We need to convert degrees latitude and longitude to radians, and we need to know the length in km of one radian on the earth's surface, which is 6378.388. The function:
set log_bin_trust_function_creators=TRUE;
DROP FUNCTION IF EXISTS GeoDistKM;
DELIMITER |
CREATE FUNCTION GeoDistKM( lat1 FLOAT, lon1 FLOAT, lat2 FLOAT, lon2 FLOAT ) RETURNS float
BEGIN
DECLARE pi, q1, q2, q3 FLOAT;
DECLARE rads FLOAT DEFAULT 0;
SET pi = PI();
SET lat1 = lat1 * pi / 180;
SET lon1 = lon1 * pi / 180;
SET lat2 = lat2 * pi / 180;
SET lon2 = lon2 * pi / 180;
SET q1 = COS(lon1-lon2);
SET q2 = COS(lat1-lat2);
SET q3 = COS(lat1+lat2);
SET rads = ACOS( 0.5*((1.0+q1)*q2 - (1.0-q1)*q3) );
RETURN 6378.388 * rads;
END;
|
DELIMITER ;
-- toronto to montreal (505km):
select geodistkm(43.6667,-79.4167,45.5000,-73.5833);
+----------------------------------------------+
| geodistkm(43.6667,-79.4167,45.5000,-73.5833) |
+----------------------------------------------+
| 505.38836669921875 |
+----------------------------------------------+
(Setting log_bin_trust_function_creators is the most convenient way to step round determinacy conventions implemented since 5.0.6.)
Moving average
Given a table of dates and daily values, retrieve their moving 5-day average:
DROP TABLE IF EXISTS t;
CREATE TABLE t (dt DATE, qty INT);
INSERT INTO t VALUES ('2007-1-1',5),
('2007-1-2',6),
('2007-1-3',7),
('2007-1-4',8),
('2007-1-5',9),
('2007-1-6',10),
('2007-1-7',11),
('2007-1-8',12),
('2007-1-9',13);
SELECT
t1.dt,
( SELECT SUM(t2.qty) / COUNT(t2.qty)
FROM t AS t2
WHERE DATEDIFF(t1.dt, t2.dt) BETWEEN 0 AND 4
) AS '5dayMovingAvg'
FROM t AS t1
ORDER BY t1.dt;
This generalises easily to moving averages for multiple items:
DROP TABLE IF EXISTS t;
CREATE TABLE t (item int, dt DATE, qty INT);
INSERT INTO t VALUES (1,'2007-1-1',5),
(1,'2007-1-2',6),
(1,'2007-1-3',7),
(1,'2007-1-4',8),
(1,'2007-1-5',9),
(1,'2007-1-6',10),
(1,'2007-1-7',11),
(1,'2007-1-8',12),
(1,'2007-1-9',13),
(2,'2007-1-1',6),
(2,'2007-1-2',7),
(2,'2007-1-3',8),
(2,'2007-1-4',9),
(2,'2007-1-5',10),
(2,'2007-1-6',11),
(2,'2007-1-7',12),
(2,'2007-1-8',13),
(2,'2007-1-9',14);
SELECT
t1.item,t1.dt,
( SELECT SUM(t2.qty) / COUNT(t2.qty)
FROM t AS t2
WHERE item=t1.item AND DATEDIFF(t1.dt, t2.dt) BETWEEN 0 AND 4
) AS '5dayMovingAvg'
FROM t AS t1
GROUP BY t1.item,t1.dt;
+------+------------+---------------+
| item | dt | 5dayMovingAvg |
+------+------------+---------------+
| 1 | 2007-01-01 | 5.0000 |
| 1 | 2007-01-02 | 5.5000 |
| 1 | 2007-01-03 | 6.0000 |
| 1 | 2007-01-04 | 6.5000 |
| 1 | 2007-01-05 | 7.0000 |
| 1 | 2007-01-06 | 8.0000 |
| 1 | 2007-01-07 | 9.0000 |
| 1 | 2007-01-08 | 10.0000 |
| 1 | 2007-01-09 | 11.0000 |
| 2 | 2007-01-01 | 6.0000 |
| 2 | 2007-01-02 | 6.5000 |
| 2 | 2007-01-03 | 7.0000 |
| 2 | 2007-01-04 | 7.5000 |
| 2 | 2007-01-05 | 8.0000 |
| 2 | 2007-01-06 | 9.0000 |
| 2 | 2007-01-07 | 10.0000 |
| 2 | 2007-01-08 | 11.0000 |
| 2 | 2007-01-09 | 12.0000 |
+------+------------+---------------+
or more simply...
SELECT t1.item,t1.dt, AVG(t2.qty) AS 5DayAvg
FROM t t1
JOIN t t2 ON t1.item=t2.item AND DATEDIFF(t1.dt, t2.dt) BETWEEN 0 AND 4
GROUP BY t1.item,t1.dt;
Multiple sums across a join
You have a parties table that holds info on peoples' names etc, and a contracts table in which each row defines one contract, identifying a client as clientpartyID and a contractor as contractorpartyID, each of these a foreign key referencing parties.partyID. You want a list of parties showing how many contracts they have participated in as client, and how many they've participated in as contractor.
SELECT
p.partyID,
p.name,
(SELECT COUNT(*) FROM contractor_client c1 WHERE c1.clientpartyID = p.partyID )
AS ClientDeals,
(SELECT COUNT(*) FROM contractor_client c2 WHERE c2.contractorpartyID = p.partyID)
AS ContractorDeals
FROM parties p
ORDER BY partyID;
Percentiles
In the Sakila table
film, retrieve a top-down percentile ranking of film lengths:
SELECT
a.film_id ,
ROUND( 100.0 * ( SELECT COUNT(*) FROM film AS b WHERE b.length <= a.length ) / total.cnt, 1 )
AS percentile
FROM film a
CROSS JOIN (
SELECT COUNT(*) AS cnt
FROM film
) AS total
ORDER BY percentile DESC;
If there are
NULLs, filter them out before computing percentiles.
On his blog, Roland Bouman
shows a much faster query; here is a version retrieving the first film at or above the 90th percentile:
SELECT
SUM(g1.r) sr,
g2.length l,
SUM(g1.r)/(SELECT COUNT(*) FROM film) p
FROM (
SELECT COUNT(*) r, length
FROM film
GROUP BY length
) g1
JOIN (
SELECT COUNT(*) r, length
FROM film
GROUP BY length
) g2 ON g1.length < g2.length
GROUP BY g2.length
HAVING p >= 0.9
ORDER BY p LIMIT 1
Random row selection
When your web page loads it is to provide a randomly selected Murphy's Law from your murphy table (id int, text law)):
SELECT law
FROM murphy
ORDER BY RAND()
LIMIT 1;
Running Sum
A user variable can maintain a per-row cumulative sum of column values:
SET @total=0;
SELECT id, value, @total:=@total+value AS RunningSum
FROM tbl;
If your platform does not permit multiple queries per connection, and if you can tolerate the O(N2) inefficiency of a self-join, this does the same job:
SELECT c.id, c.value, d.RunningSum
FROM tbl c
JOIN (
SELECT a.id, SUM(b.value) AS RunningSum
FROM tbl a
LEFT JOIN tbl b ON b.id <= a.id
GROUP BY a.id
) d USING (id);
Last updated 17 Jun 2009
Sum across categories
You often need to sum across several categories to total customer purchase amounts, salesperson sales amounts, political party election spending, etc.
For this example assume three tables: candidates, parties and ridings. You want to get the total amount spent in all ridings by every party in one output row. Here is the schema:
CREATE TABLE candidates (
id int(11) NOT NULL default '0',
`name` char(10) ,
riding char(12) ,
party char(12) ,
amt_spent decimal(10,0) NOT NULL default '0',
PRIMARY KEY (id)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO candidates
VALUES (1,'Anne Jones','Essex','Liberal','5000'),
(2,'Mary Smith','Malton','Liberal','7000'),
(3,'Sara Black','Riverdale','Liberal','15000'),
(4,'Paul Jones','Essex','Socialist','3000'),
(5,'Ed While','Essex','Conservative','10000'),
(6,'Jim kelly','Malton','Liberal','9000'),
(7,'Fred Price','Riverdale','Socialist','4000');
CREATE TABLE ridings (
riding char(10) NOT NULL default '',
PRIMARY KEY (riding)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO ridings VALUES ('Essex'),('Malton'),('Riverdale');
CREATE TABLE parties (
party char(12) NOT NULL default '',
PRIMARY KEY (party)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO parties VALUES ('Conservative'),('Liberal'),('Socialist');
And here is the query that does it:
SELECT
SUM(amt_spent) AS Total,
(SUM(amt_spent)-SUM(CASE WHEN data.party='Conservative' THEN NULL ELSE amt_spent END)) AS Cons,
(SUM(amt_spent)-SUM(CASE WHEN data.party='Liberal' THEN NULL ELSE amt_spent END)) AS Lib,
(SUM(amt_spent)-SUM(CASE WHEN data.party='Socialist' THEN NULL ELSE amt_spent END)) AS Soc
FROM
(SELECT * FROM candidates
INNER JOIN parties ON candidates.party=parties.party
INNER JOIN ridings ON candidates.riding=ridings.riding) AS data
------- ------- ------- ------
| Total | Cons | Lib | Soc |
------- ------- ------- ------
| 53000 | 10000 | 36000 | 7000 |
------- ------- ------- ------
Top ten
We often want to know the top 1, 2, 10 or whatever values from a query. This is dead simple in MySQL. However many JOINs and WHEREs the query has, simply ORDER BY the column(s) whose highest values are sought, and LIMIT the resultset:
SELECT (somecolumn), (othercolumns) ...
FROM (some tables) ...
ORDER BY somecolumn DESC
LIMIT 10;
A cursor if necessary, but not necessarily a cursor
You have photos (id INT, photo BLOB, tally INT) and votes(id INT, userID INT, photoID INT) tables. You wish to update photos.tally values from counts per photo in the votes table. You can use a cursor to walk the photos table, updating the tally as you go:
DROP TABLE IF EXISTS photos;
CREATE TABLE photos (id INT, photo BLOB, tally INT);
INSERT INTO photos VALUES(1,'',0),(2,'',0);
DROP TABLE IF EXISTS VOTES;
CREATE TABLE VOTES( userID INT, photoID INT);
INSERT INTO votes VALUES (1,1),(2,1),(2,2);
DROP PROCEDURE IF EXISTS updatetallies;
DELIMITER //
CREATE PROCEDURE updatetallies()
BEGIN
DECLARE done INT DEFAULT 0;
DECLARE pid INT;
DECLARE cur1 CURSOR FOR SELECT id FROM photos;
DECLARE CONTINUE HANDLER FOR SQLSTATE '02000' SET done = 1;
OPEN cur1;
FETCH cur1 INTO pid;
WHILE done = 0 DO
UPDATE photos
SET tally = (SELECT COUNT(*) FROM votes WHERE photoid = pid )
WHERE id = pid;
FETCH cur1 INTO pid;
END WHILE;
CLOSE cur1;
SELECT id,tally FROM photos;
END //
DELIMITER ;
CALL updatetallies();
+------+-------+
| id | tally |
+------+-------+
| 1 | 2 |
| 2 | 1 |
+------+-------+
but a simple join does exactly the same job at much less cost:
UPDATE photos
SET tally = (
SELECT COUNT(*) FROM votes WHERE votes.photoid = photos.id
);
Before you burden your app with a cursor, see if you can simplify the processing to a straightforward join.
Emulate sp_exec
Sometimes it is desirable to call multiple stored procedures in one command. In SQL Server this can be done with sp_exec. In MySQL we can easily write such an sproc that calls as many sprocs as we please, for example...
USE sys;
DROP PROCEDURE IF EXISTS sp_exec;
DELIMITER |
CREATE PROCEDURE sp_exec( p1 CHAR(64), p2 CHAR(64) )
BEGIN
-- permit doublequotes to delimit data
SET @sqlmode=(SELECT @@sql_mode);
SET @@sql_mode='';
SET @sql = CONCAT( "CALL ", p1 );
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
SET @sql = CONCAT( "CALL ", p2 );
PREPARE stmt FROM @sql;
EXECUTE stmt;
DROP PREPARE stmt;
SET @@sql_mode=@sqlmode;
END;
|
DELIMITER ;
Variable-length argument for query IN() clause
To have an sproc accept a variable-length parameter list for an IN(...) clause in a query, code the sproc to PREPARE the query statement:
DROP PROCEDURE IF EXISTS passInParam;
DELIMITER |
CREATE PROCEDURE passInParam( IN qry VARCHAR(100), IN param VARCHAR(1000) )
BEGIN
SET @qry = CONCAT( qry, param, ')' );
PREPARE stmt FROM @qry;
EXECUTE stmt;
DROP PREPARE stmt;
END;
|
DELIMITER ;
For this example, the query string should be of the form:
SELECT ... FROM ... WHERE ... IN (
but so long as it has those elements, it can be as complex as you like. When you call the sproc:
1. Quote each argument with a pair of single quotes,
2. Separate these quoted arguments with commas,
3. Surround the whole param string with another set of single quotes:
CALL passInParam( 'SELECT * FROM tbl WHERE colval IN (', ('''abc'',''def'',''ghi''' ));
Count delimited substrings
Here is a function to count substrings delimited by a constant delimiting string:
DROP FUNCTION IF EXISTS strcount;
SET GLOBAL log_bin_trust_function_creators=1;
DELIMITER |
CREATE FUNCTION strCount( pDelim VARCHAR(32), pStr TEXT) RETURNS int(11)
BEGIN
DECLARE n INT DEFAULT 0;
DECLARE pos INT DEFAULT 1;
DECLARE strRemain TEXT;
SET strRemain = pStr;
SET pos = LOCATE( pDelim, strRemain );
WHILE pos != 0 DO
SET n = n + 1;
SET pos = LOCATE( pDelim, strRemain );
SET strRemain = SUBSTRING( strRemain, pos+1 );
END WHILE;
RETURN n;
END |
DELIMITER ;
-- example call:
SET @str = "The quick brown fox jumped over the lazy dog";
SET @delim = " ";
SELECT strCount(@delim,@str);
Count substrings
To count instances of a search string in a target string ...
- in the target string, replace the search string with a single character,
- subtract the length of the modified target string from the length of the original target string,
- divide that by the length of the search string:
SET @str = "The quick brown fox jumped over the lazy dog";
SET @find = "the";
SELECT ROUND(((LENGTH(@str) - LENGTH(REPLACE(LCASE(@str), @find, '')))/LENGTH(@find)),0)
AS COUNT;
+-------+
| COUNT |
+-------+
| 2 |
+-------+
Note that REPLACE() does a case-sensitive search; to get a case-insensitive result you must coerce target and search strings to one case.
To remove decimals from the result:
SELECT CAST((LENGTH(@str) - LENGTH(REPLACE(LCASE(@str)), @find, '')))/LENGTH(@find) AS SIGNED) AS COUNT;
Levenshtein distance
The Levenshtein distance between two strings is the minimum number of operations needed to transform one string into the other, where an operation may be insertion, deletion or substitution of one character. Jason Rust published this MySQL algorithm for it at
http://www.codejanitor.com/wp/.
CREATE FUNCTION levenshtein( s1 VARCHAR(255), s2 VARCHAR(255) )
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT;
DECLARE s1_char CHAR;
-- max strlen=255
DECLARE cv0, cv1 VARBINARY(256);
SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0;
IF s1 = s2 THEN
RETURN 0;
ELSEIF s1_len = 0 THEN
RETURN s2_len;
ELSEIF s2_len = 0 THEN
RETURN s1_len;
ELSE
WHILE j <= s2_len DO
SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1;
END WHILE;
WHILE i <= s1_len DO
SET s1_char = SUBSTRING(s1, i, 1), c = i, cv0 = UNHEX(HEX(i)), j = 1;
WHILE j <= s2_len DO
SET c = c + 1;
IF s1_char = SUBSTRING(s2, j, 1) THEN
SET cost = 0; ELSE SET cost = 1;
END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost;
IF c > c_temp THEN SET c = c_temp; END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1;
IF c > c_temp THEN
SET c = c_temp;
END IF;
SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1;
END WHILE;
SET cv1 = cv0, i = i + 1;
END WHILE;
END IF;
RETURN c;
END;
Helper function:
CREATE FUNCTION levenshtein_ratio( s1 VARCHAR(255), s2 VARCHAR(255) )
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, max_len INT;
SET s1_len = LENGTH(s1), s2_len = LENGTH(s2);
IF s1_len > s2_len THEN
SET max_len = s1_len;
ELSE
SET max_len = s2_len;
END IF;
RETURN ROUND((1 - LEVENSHTEIN(s1, s2) / max_len) * 100);
END;
Proper case
The basic idea is...
- lower-case the string
- upper-case the first character if it is a-z, and any other a-z character that follows a punctuation character
Here is the function. To make it work with strings long than 128 characters, change its input and return declarations accordingly:
DROP FUNCTION IF EXISTS proper;
SET GLOBAL log_bin_trust_function_creators=TRUE;
DELIMITER |
CREATE FUNCTION proper( str VARCHAR(128) )
RETURNS VARCHAR(128)
BEGIN
DECLARE c CHAR(1);
DECLARE s VARCHAR(128);
DECLARE i INT DEFAULT 1;
DECLARE bool INT DEFAULT 1;
DECLARE punct CHAR(18) DEFAULT ' ()[]{},.-_!@;:?/'; -- David Rabby added \\'
SET s = LCASE( str );
WHILE i <= LENGTH( str ) DO -- Jesse Palmer corrected from < to <= for last char
BEGIN
SET c = SUBSTRING( s, i, 1 );
IF LOCATE( c, punct ) > 0 THEN
SET bool = 1;
ELSEIF bool=1 THEN
BEGIN
IF c >= 'a' AND c <= 'z' THEN
BEGIN
SET s = CONCAT(LEFT(s,i-1),UCASE(c),SUBSTRING(s,i+1));
SET bool = 0;
END;
ELSEIF c >= '0' AND c <= '9' THEN
SET bool = 0;
END IF;
END;
END IF;
SET i = i+1;
END;
END WHILE;
RETURN s;
END;
|
DELIMITER ;
select proper("d'arcy");
+------------------+
| proper("d'arcy") |
+------------------+
| D'Arcy |
+------------------+
But there are always exceptions, for example some guy with that name will want it spelled "d'Arcy".
Retrieve octets from IP addresses
If ip is for the form N.N.N.N where N may be 1, 2 or 3 digits, how to group and count by just the first three octets, ie the ip class?
SELECT
LEFT(ip, CHAR_LENGTH(ip) - LOCATE('.', REVERSE(ip))) as ipclass,
COUNT(*)
FROM tbl
GROUP BY ipclass;
Hamilton Turner notes we can find the first octet with LEFT(ip,LOCATE('.',ip)-1).
Return digits or alphas from a string
Return only the digits from a string:
SET GLOBAL log_bin_trust_function_creators=1;
DROP FUNCTION IF EXISTS digits;
DELIMITER |
CREATE FUNCTION digits( str CHAR(32) ) RETURNS CHAR(32)
BEGIN
DECLARE i, len SMALLINT DEFAULT 1;
DECLARE ret CHAR(32) DEFAULT '';
DECLARE c CHAR(1);
SET len = CHAR_LENGTH( str );
REPEAT
BEGIN
SET c = MID( str, i, 1 );
IF c BETWEEN '0' AND '9' THEN
SET ret=CONCAT(ret,c);
END IF;
SET i = i + 1;
END;
UNTIL i > len END REPEAT;
RETURN ret;
END |
DELIMITER ;
SELECT digits('123ab45cde6789fg');
+----------------------------+
| digits('123ab45cde6789fg') |
+----------------------------+
| 123456789 |
+----------------------------+
Return only the alphabetic characters from a string:
DROP FUNCTION IF EXISTS alphas;
DELIMITER |
CREATE FUNCTION alphas( str CHAR(32) ) RETURNS CHAR(16)
BEGIN
DECLARE i, len SMALLINT DEFAULT 1;
DECLARE ret CHAR(32) DEFAULT '';
DECLARE c CHAR(1);
SET len = CHAR_LENGTH( str );
REPEAT
BEGIN
SET c = MID( str, i, 1 );
IF c REGEXP '[[:alpha:]]' THEN
SET ret=CONCAT(ret,c);
END IF;
SET i = i + 1;
END;
UNTIL i > len END REPEAT;
RETURN ret;
END |
DELIMITER ;
SELECT alphas('123ab45cde6789fg');
+----------------------------+
| alphas('123ab45cde6789fg') |
+----------------------------+
| abcdefg |
+----------------------------+
Strip HTML tags
Ported from a T-SQL function by Robert Davis:
SET GLOBAL log_bin_trust_function_creators=1;
DROP FUNCTION IF EXISTS fnStripTags;
DELIMITER |
CREATE FUNCTION fnStripTags( Dirty varchar(4000) )
RETURNS varchar(4000)
DETERMINISTIC
BEGIN
DECLARE iStart, iEnd, iLength int;
WHILE Locate( '<', Dirty ) > 0 And Locate( '>', Dirty, Locate( '<', Dirty )) > 0 DO
BEGIN
SET iStart = Locate( '<', Dirty ), iEnd = Locate( '>', Dirty, Locate('<', Dirty ));
SET iLength = ( iEnd - iStart) + 1;
IF iLength > 0 THEN
BEGIN
SET Dirty = Insert( Dirty, iStart, iLength, '');
END;
END IF;
END;
END WHILE;
RETURN Dirty;
END;
|
DELIMITER ;
SELECT fnStripTags('this is a test, nothing more') AS Test;
+------------------------------+
| Test |
+------------------------------+
| this is a test, nothing more |
+------------------------------+