LeetCode:Database 55.报告的记录 II
Posted Xiao Miao
tags:
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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;
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