文章目录
- 1. 题目
- 2. 解题
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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