MySQL中MAX函数与Group By一起使用的注意事项

 mysql> select * from test;
 +----+-------+------+-------+
 | id | name  | age  | class |
 +----+-------+------+-------+
 |  1 | qiu   |   22 |     1 |
 |  2 | liu   |   42 |     1 |
 |  4 | zheng |   20 |     2 |
 |  3 | qian  |   20 |     2 |
 |  0 | wang  |   11 |     3 |
 |  6 | li    |   33 |     3 |
 +----+-------+------+-------+
 6 rows in set (0.00 sec)

如果想找到每个class里面的最大的age，则需要使用group by和max。

如下的sql语句，则输出结果有错误：

 mysql> select id,name,max(age),class from test group by class;
 +----+-------+----------+-------+
 | id | name  | max(age) | class |
 +----+-------+----------+-------+
 |  1 | qiu   |       42 |     1 |
 |  4 | zheng |       20 |     2 |
 |  0 | wang  |       33 |     3 |
 +----+-------+----------+-------+
 3 rows in set (0.00 sec)

虽然找到的age是最大的age，但是与之匹配的用户信息却不是真实的信息，而是group by分组后的第一条记录的基本信息。

如果我使用以下的语句进行查找，则可以返回真实的结果。

 mysql> select * from (
     -> select * from test order by age desc) as b
     -> group by class;
 +----+-------+------+-------+
 | id | name  | age  | class |
 +----+-------+------+-------+
 |  2 | liu   |   42 |     1 |
 |  4 | zheng |   20 |     2 |
 |  6 | li    |   33 |     3 |
 +----+-------+------+-------+
 3 rows in set (0.00 sec)

方法2：

    

 select * from test t where t.age = (select max(age) from test where t.class = class) order by class;