LeetCode MySQL 1369. 获取最近第二次的活动（over窗口函数）

文章目录

- 1. 题目
- 2. 解题

1. 题目

表: UserActivity

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| username      | varchar |
| activity      | varchar |
| startDate     | Date    |
| endDate       | Date    |
+---------------+---------+
该表不包含主键
该表包含每个用户在一段时间内进行的活动的信息
名为 username 的用户在 startDate 到 endDate 日内有一次活动

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写一条SQL查询展示每一位用户最近第二次的活动（倒数第二次）

如果用户仅有一次活动，返回该活动

一个用户不能同时进行超过一项活动，以任意顺序返回结果

下面是查询结果格式的例子：

UserActivity 表:
+------------+--------------+-------------+-------------+
| username   | activity     | startDate   | endDate     |
+------------+--------------+-------------+-------------+
| Alice      | Travel       | 2020-02-12  | 2020-02-20  |
| Alice      | Dancing      | 2020-02-21  | 2020-02-23  |
| Alice      | Travel       | 2020-02-24  | 2020-02-28  |
| Bob        | Travel       | 2020-02-11  | 2020-02-18  |
+------------+--------------+-------------+-------------+

Result 表:
+------------+--------------+-------------+-------------+
| username   | activity     | startDate   | endDate     |
+------------+--------------+-------------+-------------+
| Alice      | Dancing      | 2020-02-21  | 2020-02-23  |
| Bob        | Travel       | 2020-02-11  | 2020-02-18  |
+------------+--------------+-------------+-------------+

Alice 最近第二次的活动是从 2020-02-24 到 2020-02-28 的旅行, 
	在此之前的 2020-02-21 到 2020-02-23 她进行了舞蹈
Bob 只有一条记录，我们就取这条记录

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来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/get-the-second-most-recent-activity

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

先选出只有一次活动的人的记录

select *
from UserActivity
group by username
having count(*)=1

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{"headers": ["username", "activity", "startDate", "endDate"], 
"values": [["Bob", "Travel", "2020-02-11", "2020-02-18"]]}

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窗口函数选出每个人倒数第二次的活动

select username, activity, startDate, endDate
from
(
    select *, rank() over(partition by username order by startDate desc) rnk
    from UserActivity
) t
where rnk = 2

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{"headers": ["username", "activity", "startDate", "endDate"], 
"values": [["Alice", "Dancing", "2020-02-21", "2020-02-23"]]}

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最后合并

# Write your MySQL query statement below
select *
from
(
    select *
    from UserActivity
    group by username
    having count(*)=1

    union all
    
    select username, activity, startDate, endDate
    from
    (
        select *, rank() over(partition by username order by startDate desc) rnk
        from UserActivity
    ) t
    where rnk = 2
) t

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更简洁的写法

# Write your MySQL query statement below
select username, activity, startDate, endDate
from
(
    select *,
        rank() over (partition by username order by startDate desc) rnk,
        count(*) over (partition by username) cnt
    from UserActivity
) t
where cnt = 1 or rnk = 2

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