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题解 | #每篇文章同一时刻最大在看人数#
SELECT artical_id, MAX(total_num) max_uv FROM (SELECT artical_id, SUM(num) over (partition by artical_id order by dt, num DESC) total_num FROM (SELEC...
Mysql
2021-12-12
1
1319
题解 | #国庆期间每类视频点赞量和转发量#
SELECT * FROM( SELECT tag, dt, SUM(like_cnt) over (partition by tag order by dt rows 6 preceding) sum_like_cnt_7d, MAX(retweet_cnt) over ...
Mysql
2021-12-11
13
598
题解 | #商品交易(网易校招笔试真题)#
法一:(窗口函数) SELECT DISTINCT id,name,weight,total FROM (SELECT goods.id,name,weight,SUM(count) over (partition by trans.goods_id) total FROM goods JOIN...
Mysql
2021-12-10
0
423
题解 | #最差是第几名(二)#
法一: WITH t1 AS ( SELECT grade, (SUM(number) over (ORDER BY grade))-(number-1) AS min_rank, SUM(number) over (ORDER BY grade) AS max_rank ...
Mysql
2021-12-10
0
422
题解 | #牛客的课程订单分析(五)#
法一:最直接的就是借鉴#牛客的课程订单分析(四)#的做法,取rk=1的表和取rk=2的表内连接起来。好处是直观,但是代码有些冗余,就不贴了。 法二:参考评论区,我想到了IF(rk=1, date, NULL)但是会出现多个user_id和NULL的date, 没想到加上一个MAX() GROUP B...
Mysql
2021-12-09
0
340
题解 | #考试分数(五)#
法一: SELECT * FROM (SELECT *, ROW_NUMBER() over (partition by job order by score DESC) t_rank FROM grade) t1 WHERE (job,t_rank) IN (SELECT job,t_medium...
Mysql
2021-12-09
0
360
题解 | #考试分数(四)#
法一(先窗口函数排序了,其实不需要的,因为只需要分组后的总数就可以了,捂脸): SELECT job, case when MAX(rk)%2=0 then ROUND(MAX(rk)/2) else ROUND(MAX(rk)/2) end AS start, case when MAX(rk)%...
Mysql
2021-12-09
0
374
题解 | #牛客每个人最近的登录日期(五)#
方法一: WITH t1 AS( SELECT date,COUNT(user_id) retention_num FROM login WHERE (user_id,date) IN (SELECT user_id,DATE_ADD(MIN(date),INTERVAL 1 DAY) ...
Mysql
2021-12-08
1
687
题解 | #牛客每个人最近的登录日期(四)#
这题和这个https://blog.nowcoder.net/n/790154f3ad084dcfbf6042b42a8dc99f 找新增用户数很像。 本打算如下。但是下面的代码不会统计出新增用户数为0的日期,和题目的答案不相符。 SELECT date, COUNT(user_id) new FR...
Mysql
2021-12-07
0
442
题解 | #牛客每个人最近的登录日期(三)#
法一: SELECT ROUND(COUNT(t2.user_id)/ COUNT(t1.user_id),3) p FROM (SELECT user_id, MIN(date) date FROM login GROUP BY user_id) t1 LEFT JOIN (SELECT DIS...
Mysql
2021-12-07
4
749
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