Here are examples of how to solve some common problems with MySQL.
Some of the examples use the table shop to hold the price of each article (item number) for certain traders (dealers). Supposing that each trader has a single fixed price per article, then (article, dealer) is a primary key for the records.
Start the command-line tool mysql and select a database:
mysql your-database-name
(In most MySQL installations, you can use the database-name ‘test').
You can create the example table as:
CREATE TABLE shop ( article INT(4) UNSIGNED ZEROFILL DEFAULT '0000' NOT NULL, dealer CHAR(20) DEFAULT '' NOT NULL, price DOUBLE(16,2) DEFAULT '0.00' NOT NULL, PRIMARY KEY(article, dealer)); INSERT INTO shop VALUES (1,'A',3.45),(1,'B',3.99),(2,'A',10.99),(3,'B',1.45),(3,'C',1.69), (3,'D',1.25),(4,'D',19.95);
Okay, so the example data is:
mysql> SELECT * FROM shop; +---------+--------+-------+ | article | dealer | price | +---------+--------+-------+ | 0001 | A | 3.45 | | 0001 | B | 3.99 | | 0002 | A | 10.99 | | 0003 | B | 1.45 | | 0003 | C | 1.69 | | 0003 | D | 1.25 | | 0004 | D | 19.95 | +---------+--------+-------+
“What’s the highest item number?”
SELECT MAX(article) AS article FROM shop +---------+ | article | +---------+ | 4 | +---------+
“Find number, dealer, and price of the most expensive article.”
In ANSI SQL this is easily done with a sub-query:
SELECT article, dealer, price FROM shop WHERE price=(SELECT MAX(price) FROM shop)
In MySQL (which does not yet have sub-selects), just do it in two steps:
Get the maximum price value from the table with a SELECT statement.
Using this value compile the actual query:
SELECT article, dealer, price FROM shop WHERE price=19.95
Another solution is to sort all rows descending by price and only get the first row using the MySQL-specific LIMIT clause:
SELECT article, dealer, price FROM shop ORDER BY price DESC LIMIT 1
NOTE: If there are several most expensive articles (for example, each $19.95) the LIMIT solution shows only one of them!
“What’s the highest price per article?”
SELECT article, MAX(price) AS price FROM shop GROUP BY article +---------+-------+ | article | price | +---------+-------+ | 0001 | 3.99 | | 0002 | 10.99 | | 0003 | 1.69 | | 0004 | 19.95 | +---------+-------+
“For each article, find the dealer(s) with the most expensive price.”
In ANSI SQL, I’d do it with a sub-query like this:
SELECT article, dealer, price
FROM shop s1
WHERE price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article);In MySQL it’s best to do it in several steps:
Get the list of (article,maxprice).
For each article get the corresponding rows that have the stored maximum price.
This can easily be done with a temporary table:
CREATE TEMPORARY TABLE tmp (
article INT(4) UNSIGNED ZEROFILL DEFAULT '0000' NOT NULL,
price DOUBLE(16,2) DEFAULT '0.00' NOT NULL);
LOCK TABLES shop read;
INSERT INTO tmp SELECT article, MAX(price) FROM shop GROUP BY article;
SELECT shop.article, dealer, shop.price FROM shop, tmp
WHERE shop.article=tmp.article AND shop.price=tmp.price;
UNLOCK TABLES;
DROP TABLE tmp;If you don’t use a TEMPORARY table, you must also lock the ‘tmp’ table. “Can it be done with a single query?” Yes, but only by using a quite inefficient trick that I call the “MAX-CONCAT trick”:
SELECT article,
SUBSTRING( MAX( CONCAT(LPAD(price,6,'0'),dealer) ), 7) AS dealer,
0.00+LEFT( MAX( CONCAT(LPAD(price,6,'0'),dealer) ), 6) AS price
FROM shop
GROUP BY article;
+---------+--------+-------+
| article | dealer | price |
+---------+--------+-------+
| 0001 | B | 3.99 |
| 0002 | A | 10.99 |
| 0003 | C | 1.69 |
| 0004 | D | 19.95 |
+---------+--------+-------+The last example can, of course, be made a bit more efficient by doing the splitting of the concatenated column in the client.
You can use MySQL user variables to remember results without having to store them in temporary variables in the client. See Section 6.1.4.
For example, to find the articles with the highest and lowest price you can do:
mysql> SELECT @min_price:=MIN(price),@max_price:=MAX(price) FROM shop; mysql> SELECT * FROM shop WHERE price=@min_price OR price=@max_price; +---------+--------+-------+ | article | dealer | price | +---------+--------+-------+ | 0003 | D | 1.25 | | 0004 | D | 19.95 | +---------+--------+-------+
In MySQL 3.23.44 and up, InnoDB tables support checking of foreign key constraints. See Section 7.5. See also Section 1.7.4.5.
You don’t actually need foreign keys to join 2 tables. The only things MySQL currently doesn’t do (in types other than InnoDB), are CHECK to make sure that the keys you use really exist in the table(s) you’re referencing and automatically delete rows from a table with a foreign key definition. If you use your keys like normal, it’ll work just fine:
CREATE TABLE person (
id SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
name CHAR(60) NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE shirt (
id SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
style ENUM('t-shirt', 'polo', 'dress') NOT NULL,
color ENUM('red', 'blue', 'orange', 'white', 'black') NOT NULL,
owner SMALLINT UNSIGNED NOT NULL REFERENCES persons,
PRIMARY KEY (id)
);
INSERT INTO person VALUES (NULL, 'Antonio Paz');
INSERT INTO shirt VALUES
(NULL, 'polo', 'blue', LAST_INSERT_ID( )),
(NULL, 'dress', 'white', LAST_INSERT_ID( )),
(NULL, 't-shirt', 'blue', LAST_INSERT_ID( ));
INSERT INTO person VALUES (NULL, 'Lilliana Angelovska');
INSERT INTO shirt VALUES
(NULL, 'dress', 'orange', LAST_INSERT_ID( )),
(NULL, 'polo', 'red', LAST_INSERT_ID( )),
(NULL, 'dress', 'blue', LAST_INSERT_ID( )),
(NULL, 't-shirt', 'white', LAST_INSERT_ID( ));
SELECT * FROM person;
+----+---------------------+
| id | name |
+----+---------------------+
| 1 | Antonio Paz |
| 2 | Lilliana Angelovska |
+----+---------------------+
SELECT * FROM shirt;
+----+---------+--------+-------+
| id | style | color | owner |
+----+---------+--------+-------+
| 1 | polo | blue | 1 |
| 2 | dress | white | 1 |
| 3 | t-shirt | blue | 1 |
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
| 7 | t-shirt | white | 2 |
+----+---------+--------+-------+
SELECT s.* FROM person p, shirt s
WHERE p.name LIKE 'Lilliana%'
AND s.owner = p.id
AND s.color <> 'white';
+----+-------+--------+-------+
| id | style | color | owner |
+----+-------+--------+-------+
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
+----+-------+--------+-------+MySQL doesn’t yet optimise when you search on two different keys combined with OR (searching on one key with different OR parts is optimised quite well):
SELECT field1_index, field2_index FROM test_table WHERE field1_index = '1' OR field2_index = '1'
The reason is that we haven’t yet had time to come up with an efficient way to handle this in the general case. (The AND handling is, in comparison, now completely general and works very well.)
For the moment you can solve this very efficiently by using a TEMPORARY table. This type of optimisation is also very good if you are using very complicated queries where the SQL server does the optimisations in the wrong order.
CREATE TEMPORARY TABLE tmp SELECT field1_index, field2_index FROM test_table WHERE field1_index = '1'; INSERT INTO tmp SELECT field1_index, field2_index FROM test_table WHERE field2_index = '1'; SELECT * from tmp; DROP TABLE tmp;
The way to solve the query shown here is, in effect, a UNION of two queries. See Section 6.4.1.2.
The following shows an idea of how you can use the bit group functions to calculate the number of days per month a user has visited a web page:
CREATE TABLE t1 (year YEAR(4), month INT(2) UNSIGNED ZEROFILL,
day INT(2) UNSIGNED ZEROFILL);
INSERT INTO t1 VALUES(2000,1,1),(2000,1,20),(2000,1,30),(2000,2,2),
(2000,2,23),(2000,2,23);
SELECT year,month,BIT_COUNT(BIT_OR(1<<day)) AS days FROM t1
GROUP BY year,month;
Which returns:
+------+-------+------+
| year | month | days |
+------+-------+------+
| 2000 | 01 | 3 |
| 2000 | 02 | 2 |
+------+-------+------+The above calculates how many different days was used for a given year/month combination, with automatic removal of duplicate entries.
The AUTO_INCREMENT attribute can be used to generate a unique identity for new rows:
CREATE TABLE animals (id MEDIUMINT NOT NULL AUTO_INCREMENT,
name CHAR(30) NOT NULL, PRIMARY KEY (id));
INSERT INTO animals (name) VALUES ("dog"),("cat"),("penguin"),
("lax"),("whale");
SELECT * FROM animals;
Which returns:
+----+---------+
| id | name |
+----+---------+
| 1 | dog |
| 2 | cat |
| 3 | penguin |
| 4 | lax |
| 5 | whale |
+----+---------+For MyISAM and BDB tables you can specify AUTO_INCREMENT on secondary column in a multi-column key. In this case the generated value for the auto-increment column is calculated as MAX(auto_increment_column)+1) WHERE prefix=given-prefix. This is useful when you want to put data into ordered groups.
CREATE TABLE animals (grp ENUM('fish','mammal','bird') NOT NULL,
id MEDIUMINT NOT NULL AUTO_INCREMENT
PRIMARY KEY (grp,id));
INSERT INTO animals (grp,name) VALUES("mammal","dog"),("mammal","cat"),
("bird","penguin"),("fish","lax"),("mammal","whale");
SELECT * FROM animals ORDER BY grp,id;
Which returns:
+--------+----+---------+
| grp | id | name |
+--------+----+---------+
| fish | 1 | lax |
| mammal | 1 | dog |
| mammal | 2 | cat |
| mammal | 3 | whale |
| bird | 1 | penguin |
+--------+----+---------+Note that in this case, the AUTO_INCREMENT value will be reused if you delete the row with the biggest AUTO_INCREMENT value in any group.
You can get the used AUTO_INCREMENT key with the LAST_INSERT_ID( ) SQL function or the mysql_insert_id( ) API function.
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