LeetCode：Database 55.报告的记录 II

Posted 2021-06-22 Xiao Miao

要求：编写一段 SQL 来查找：在被报告为垃圾广告的帖子中，被移除的帖子的每日平均占比，四舍五入到小数点后2位。

Actions 表的结构：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| post_id       | int     |
| action_date   | date    |
| action        | enum    |
| extra         | varchar |
+---------------+---------+
这张表没有主键，并有可能存在重复的行。

Removals 表的结构：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| post_id       | int     |
| remove_date   | date    | 
+---------------+---------+
这张表的主键是 post_id。

Actions 表:

+---------+---------+-------------+--------+--------+
| user_id | post_id | action_date | action | extra  |
+---------+---------+-------------+--------+--------+
| 1       | 1       | 2019-07-01  | view   | null   |
| 1       | 1       | 2019-07-01  | like   | null   |
| 1       | 1       | 2019-07-01  | share  | null   |
| 2       | 2       | 2019-07-04  | view   | null   |
| 2       | 2       | 2019-07-04  | report | spam   |
| 3       | 4       | 2019-07-04  | view   | null   |
| 3       | 4       | 2019-07-04  | report | spam   |
| 4       | 3       | 2019-07-02  | view   | null   |
| 4       | 3       | 2019-07-02  | report | spam   |
| 5       | 2       | 2019-07-03  | view   | null   |
| 5       | 2       | 2019-07-03  | report | racism |
| 5       | 5       | 2019-07-03  | view   | null   |
| 5       | 5       | 2019-07-03  | report | racism |
+---------+---------+-------------+--------+--------+

Removals 表:

+---------+-------------+
| post_id | remove_date |
+---------+-------------+
| 2       | 2019-07-20  |
| 3       | 2019-07-18  |
+---------+-------------+

Result Table:

+-----------------------+
| average_daily_percent |
+-----------------------+
| 75.00                 |
+-----------------------+
2019-07-04 的垃圾广告移除率是 50%，因为有两张帖子被报告为垃圾广告，但只有一个得到移除。
2019-07-02 的垃圾广告移除率是 100%，因为有一张帖子被举报为垃圾广告并得到移除。
其余几天没有收到垃圾广告的举报，因此平均值为：(50 + 100) / 2 = 75%
注意，输出仅需要一个平均值即可，我们并不关注移除操作的日期。

SQL语句：
注：on对左外的表和右外的表无效，需要使用where过滤

#1.方法1
SELECT ROUND(AVG(c1)*100,2) AS average_daily_percent
FROM( 
SELECT COUNT(IFNULL(DISTINCT b.post_id,0))/COUNT(DISTINCT a.post_id) AS c1,a.action_date AS d1
FROM a  
JOIN a b
ON a.action_date=b.action_date AND b.post_id IN(SELECT post_id FROM removals)  
GROUP BY a.action_date)c

#2.方法2
select round(avg(c1)*100,2) as average_daily_percent
from(
select count(distinct a2.post_id)/count(distinct a1.post_id) as c1
from actions as a1
left join actions as a2
on a1.action_date=a2.action_date
and a2.post_id in(select post_id from removals)
where a1.extra='spam'
group by a1.action_date)b;