哪个司机会开所有类型的车？这便是一个关系除法问题。

##创建表 表drivers有两个字段，司机的名字和司机会开的车的id：

CREATE TABLE `drivers` (

`driver_name` char(10) DEFAULT NULL,

`vehicle_id` int(11) DEFAULT NULL

) ENGINE=InnoDB DEFAULT CHARSET=utf8;

注意，实际情况下会有更多的列，且driver_name和vehicle_id应该共同作为主键。

在表drivers中插入以下数据：

+-------------+------------+

| driver_name | vehicle_id |

+-------------+------------+

| yan | 1 |

| wei | 1 |

| li | 1 |

| wei | 2 |

| wei | 3 |

| li | 2 |

+-------------+------------+

表vehicles只有一个字段，即车的id：

CREATE TABLE `vehicles` (

`id` int(11) DEFAULT NULL

) ENGINE=InnoDB DEFAULT CHARSET=utf8;

注意，实际情况下会有更多的列，且id应该作为主键。

在表vehicles中插入以下数据：

+----+

| id |

+----+

| 1 |

| 2 |

| 3 |

+----+

##方法1

SELECT DISTINCT D1.driver_name

FROM drivers AS D1

WHERE NOT EXISTS

(SELECT * FROM vehicles AS V1

WHERE NOT EXISTS

(

SELECT * FROM drivers AS D2

WHERE D1.driver_name = D2.driver_name

AND D2.vehicle_id = V1.id

)

);

运行结果是：

+-------------+

| driver_name |

+-------------+

| wei |

+-------------+

**解释：**个人觉得这个很拗口，自己也云里雾里的。我们要找的是哪个司机会开所有的车，这也意味着，如果某个司机不会开某辆车，则这个司机不会出现在结果中。

SELECT * FROM vehicles AS V1

WHERE EXISTS

(

SELECT * FROM drivers AS D2

WHERE D1.driver_name = D2.driver_name

AND D2.vehicle_id = V1.id

)

上面的代码，认为是D1中的某个司机会开V1中的某辆车。而

SELECT * FROM vehicles AS V1

WHERE NOT EXISTS

(

SELECT * FROM drivers AS D2

WHERE D1.driver_name = D2.driver_name

AND D2.vehicle_id = V1.id

)

就可以看成是D1中的某个司机不会开V1中的某辆车。

既然这个司机不会开某辆车，那么这个司机也就不应该出现在结果中，所以使用：

SELECT DISTINCT D1.driver_name

FROM drivers AS D1

WHERE NOT EXISTS

注意，若没有DISTINCT，运行结果将是：

+-------------+

| driver_name |

+-------------+

| wei |

| wei |

| wei |

+-------------+

##方法2 这个方法比较容易理解，由于共有3辆车，所以看哪些司机会开3辆车即可：

SELECT D1.driver_name

FROM drivers AS D1, vehicles AS V1

WHERE D1.vehicle_id = V1.id

GROUP BY D1.driver_name

HAVING COUNT(D1.vehicle_id) = (SELECT COUNT(*) FROM vehicles);

运行结果是：

+-------------+

| driver_name |

+-------------+

| wei |

+-------------+

下面一步步分解：

mysql> SELECT D1.driver_name

FROM drivers AS D1, vehicles AS V1

WHERE D1.vehicle_id = V1.id;

+-------------+

| driver_name |

+-------------+

| yan |

| wei |

| li |

| wei |

| wei |

| li |

+-------------+

6 rows in set

加上GROUP后：

mysql> SELECT D1.driver_name, COUNT(*)

FROM drivers AS D1, vehicles AS V1

WHERE D1.vehicle_id = V1.id

GROUP BY D1.driver_name;

+-------------+----------+

| driver_name | COUNT(*) |

+-------------+----------+

| li | 2 |

| wei | 3 |

| yan | 1 |

+-------------+----------+

3 rows in set

使用HAVING，过滤GROUP后的结果：

mysql> SELECT D1.driver_name, COUNT(*)

FROM drivers AS D1, vehicles AS V1

WHERE D1.vehicle_id = V1.id

GROUP BY D1.driver_name

HAVING COUNT(D1.vehicle_id) = (SELECT COUNT(*) FROM vehicles);

+-------------+----------+

| driver_name | COUNT(*) |

+-------------+----------+

| wei | 3 |

+-------------+----------+

1 row in set

Joe Celko 《SQL编程风格》