以下 SQL 查询的正确解决方案是什么？

Question

我尝试了以下查询，但出现错误：

select ACTIVITY
from (select ACTIVITY, count(*) as cnt,
             max(count(*)) as max_cnt,
             min(count(*)) as min_cnt
      from FRIENDS GROUP BY ACTIVITY) FRIENDS
where cnt not in (max_cnt, min_cnt);

ERROR: ERROR 1111 (HY000) at line 1: Invalid use of group function MYSQL VERSION: 8

Answer 1

您需要 window 个功能：

select ACTIVITY
from (select ACTIVITY, count(*) as cnt,
             max(count(*)) over () as max_cnt,
             min(count(*)) over () as min_cnt
      from FRIENDS 
      group by activity
     ) a
where cnt not in (max_cnt, min_cnt);

以上要求 MySQL 8+。早期版本比较蛋疼：

select a.ACTIVITY
from (select ACTIVITY, count(*) as cnt,
             max(count(*)) over () as max_cnt,
             min(count(*)) over () as min_cnt
      from FRIENDS 
      group by activity
     ) a join
     (select min(cnt) as min_cnt, max(cnt) as max_cnt
      from (select activity, count(*) as cnt
            from friends
            group by activity
           ) a
     ) am
     on a.cnt not in (am.max_cnt, am.min_cnt);

Answer 2

尝试以下操作，它应该可以使用 MySQL 8.0 中的 window 函数。

select
    activity
from
(
    select 
        activity, 
        count(*) over () as ttl,
        dense_rank() over (order by count(*)) as rnk
    from friends 
    group by 
        activity
)  val
where rnk != 1 and rnk != ttl - 1

Answer 3

据推测，你不是运行 MySQL 8.0（否则你上一个问题给出的答案就可以了）。

在早期版本中，您可以：

select activity, count(*) no_activities
from friends
group by activity
having 
        count(*) > (select count(*) from friends group by activity order by count(*)  asc limit 1)
    and count(*) < (select count(*) from friends group by activity order by count(*) desc limit 1)

以下 SQL 查询的正确解决方案是什么？

What is the correct solution for the below SQL query?

mysql

sql

group-by

count

greatest-n-per-group