下面是一些学习如何用MySQL解决一些常见问题的例子。一些例子使用数据库表“shop”，包含某个商人的每篇文章(物品号)的价格。假定每个商人的每篇文章有一个单独的固定价格，那么(物品，商人)是记录的主键。
CREATE TABLE shop (
“最大的物品号是什么？”
“找出最贵的文章的编号、商人和价格”
“每篇文章的最高的价格是什么？”
“对每篇文章，找出有最贵的价格的交易者。”
不需要外键联结2个表。
MySQL唯一不做的事情是CHECK以保证你使用的键确实在你正在引用表中存在，并且它不自动从有一个外键定义的表中删除行。如果你象平常那样使用你的键值，它将工作得很好！
CREATE TABLE persons (
id SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
name CHAR(60) NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE shirts (
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 persons VALUES (NULL, 'Antonio Paz');
INSERT INTO shirts VALUES
(NULL, 'polo', 'blue', LAST_INSERT_ID()),
(NULL, 'dress', 'white', LAST_INSERT_ID()),
(NULL, 't-shirt', 'blue', LAST_INSERT_ID());
INSERT INTO persons VALUES (NULL, 'Lilliana Angelovska');
INSERT INTO shirts 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 persons;
+----+---------------------+
| id | name|
+----+---------------------+
| 1 | Antonio Paz |
| 2 | Lilliana Angelovska |
+----+---------------------+
SELECT * FROM shirts;
+----+---------+--------+-------+
| 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 persons p, shirts 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 |
+----+-------+--------+-------+