MySQL中MAX函数与Group By一起使用的注意事项

矫情吗；* 2022-03-16 10:40 420阅读 0赞

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;