mysql 查找重复姓名且年龄最大的列表

mysql> select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc;+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | atest  |   64 ||     2 | btest  |   37 ||     2 | ctest  |   43 ||     2 | dtest  |   43 ||     1 | mary   |   22 ||     1 | kou    |   22 ||     1 | perter |   23 ||     1 | kate   |   19 |+-------+--------+------+8 rows in set (0.00 sec)

这里找到count 重复的数据
下面接着找 count 最大，切age 最大且相同的数据

mysql> select count,name,age from ( select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc;+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | atest  |   64 ||     2 | ctest  |   43 ||     1 | perter |   23 |+-------+--------+------+3 rows in set (0.00 sec)

为什么少了一条  dtest   ，dtest的数据和ctest在count和age上是一样的？

求指教！谢谢

回复讨论(解决方案)

第二式有 group by count，那么 count 相同的肯定在一组了
既然
|     2 | ctest  |   43 |
|     2 | dtest  |   43 |
在一组，那自然只能出现一个了
所以分组条件应加上 age，即 group by count,age

mysql> select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc, ae desc ;+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | zx     |   64 ||     2 | xz     |   43 ||     2 | john   |   43 ||     2 | tom    |   37 ||     1 | perter |   23 ||     1 | mary   |   22 ||     1 | kou    |   22 ||     1 | kate   |   19 |+-------+--------+------+8 rows in set (0.00 sec)mysql> select count,name,age from ( select count(*) as count  ,name,sum(age) as age from t1 group b name order by count desc ,age desc ) as tmp group by count,age order by count desc ,age desc;+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | zx     |   64 ||     2 | xz     |   43 ||     2 | tom    |   37 ||     1 | perter |   23 ||     1 | mary   |   22 ||     1 | kate   |   19 |+-------+--------+------+6 rows in set (0.00 sec)

这个结果不对吧？

预期的结果应该是：

+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | atest  |   64 ||     2 | ctest  |   43 ||     2 | dtest  |   43 ||     1 | perter |   23 |+-------+--------+------+

select count,name,age from ( select count(*) as count  ,name,sum(age) as age from t1 group b
 name order by count desc ,age desc ) as tmp group by count, name order by count desc ,age desc;

mysql> select count,name,age from ( select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count,name order by count desc ,age desc;+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | zx     |   64 ||     2 | xz     |   43 ||     2 | john   |   43 ||     2 | tom    |   37 ||     1 | perter |   23 ||     1 | kou    |   22 ||     1 | mary   |   22 ||     1 | kate   |   19 |+-------+--------+------+8 rows in set (0.00 sec)

group by count,name 这样找不到 count=2且age最大的数据和count=1且age最大的数据了~

+-------+--------+------+| count | name   | age  |+-------+--------+------+|     3 | atest  |   64 ||     2 | ctest  |   43 ||     2 | dtest  |   43 ||     1 | perter |   23 |+-------+--------+------+

这种预期的结果，能在一条sql里体现出来么？

 

求指教

感觉效率应该不是很高，虽然可以做出来

(select count,max(age) as age from ( select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc) as tempd,( select count(*) as count  ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp1 where tmp1.count=tempd.count and tmp1.age=tempd.age count order by tempd.count desc ,tempd.age desc;

select tmp1.count,tmp1.age from (select count,max(age) as age from ( select count(*) as count  ,name,sum(age) as age from t1 group by
 name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc) as tempd,( select count(*) as count  ,name,sum(age) as age from t1 group by
 name order by count desc ,age desc ) as tmp1 where tmp1.count=tempd.count and tmp1.age=tempd.age count order by tmp1.count desc ,tmp1.age desc;

非常感谢版主