1. 题目

表: Friends

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
| activity      | varchar |
+---------------+---------+
id 是朋友的 id 和该表的主键
name 是朋友的名字
activity 是朋友参加的活动的名字

表: Activities

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是该表的主键
name 是活动的名字

写一条 SQL 查询那些既没有最多，也没有最少参与者的活动的名字

可以以任何顺序返回结果，Activities 表的每项活动的参与者都来自 Friends 表

下面是查询结果格式的例子：

Friends 表:
+------+--------------+---------------+
| id   | name         | activity      |
+------+--------------+---------------+
| 1    | Jonathan D.  | Eating        |
| 2    | Jade W.      | Singing       |
| 3    | Victor J.    | Singing       |
| 4    | Elvis Q.     | Eating        |
| 5    | Daniel A.    | Eating        |
| 6    | Bob B.       | Horse Riding  |
+------+--------------+---------------+Activities 表:
+------+--------------+
| id   | name         |
+------+--------------+
| 1    | Eating       |
| 2    | Singing      |
| 3    | Horse Riding |
+------+--------------+Result 表:
+--------------+
| activity     |
+--------------+
| Singing      |
+--------------+Eating 活动有三个人参加, 是最多人参加的活动 (Jonathan D. , Elvis Q. and Daniel A.)
Horse Riding 活动有一个人参加, 是最少人参加的活动 (Bob B.)
Singing 活动有两个人参加 (Victor J. and Jade W.)

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/activity-participants
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

# Write your MySQL query statement below
select activity
from Friends
group by activity
having count(*) > any(select count(*) from Friends group by activity)and count(*) < any(select count(*) from Friends group by activity)

or

# Write your MySQL query statement below
select activity
from Friends
group by activity
having count(*) != (select count(*) c from Friends group by activity order by c desc limit 1)and count(*) != (select count(*) c from Friends group by activity order by c limit 1)

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