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题解 | #找出待召回的流失用户# 小白拆解做法

with tiaojian as ( select uid, count(id) as times from user_login_tb group by uid having count(id)>=3 ),tiaojian1 as ( select uid, count(distinct ...

2024-02-26

0 235

题解 | #日活次日留存率和新户次日留存率# 小白复杂写法

with date as ( select uid, min(login_date) as first_date from user_login_tb group by uid ),date1 as ( select uid, login_date from user_login_tb gro...

2024-02-23

0 237

题解 | #计算每天的新用户占比#

select t.login_date as dt, count(distinct uid) as total_user_num, concat( ifnull( round( ( sum(case when t.m=1 then 1 end)/count(distinct uid))*100,1)...

2024-02-23

1 316

题解 | #实习广场投递简历分析(三)#

with tiaojian as ( select job, month(date) as months, date_format(date,"%Y-%m") as month, sum(num) as cnt from resume_info where year(date)...

2024-01-11

0 228

题解 | #牛客每个人最近的登录日期(五)#

with tiaojian as ( select date from login group by date ),tiaojian1 as ( select t.m, round( count(distinct case when datediff(t.p,t.m)=1 then t.user...

2024-01-11

0 198

题解 | #近三个月未完成试卷数为0的用户完成情况#

select t.uid, sum(case when t.pday is not null then 1 else 0 end) as exam_complete_cnt from( select uid, date(start_time) as sday, date(submit_time) ...

2023-11-30

0 263

题解 | #连续两次作答试卷的最大时间窗#

select t.uid, max(datediff(t.m,t.pday))+1 as cnt, round( (t.puid/max((datediff(t.xday,t.mday))+1))*(max(datediff(t.m,t.pday))+1),2) as pcn from( selec...

2023-11-30

0 356

题解 | #查看不同年龄段的用户明细#

select device_id, gender, case when age<20 then "20岁以下" when age between 20 and 24 then "20-24岁" when age>=25 then...

2023-11-27

1 244

题解 | #计算用户8月每天的练题数量#

select day(date), count(id) from question_practice_detail where date_format(date,"%Y%m")=202108 group by date

2023-11-27

1 230

题解 | #计算用户的平均次日留存率#

select sum(case when datediff(t.py,t.date)=1 then 1 else 0 end)/ count(distinct device_id,date) from( select id,device_id,date, lead(date,1)over(part...

2023-11-27

1 219