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;