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题解 | #牛客每个人最近的登录日期(六)#

sql script select u.name as u_n, tmp.date as date, sum(tmp.ps_num) over(partition by u.name order by date) as ps_num from (select pn.user_id as user_i...

sql

2021-09-10

0 281

题解 | #牛客每个人最近的登录日期(五)#

sql script select u.name as u_n, tmp.date as date, sum(tmp.ps_num) over(partition by u.name order by date) as ps_num from (select pn.user_id as user_i...

sql

2021-09-10

0 321

题解 | #牛客每个人最近的登录日期(五)#

sql script -- new user every day numbers select date, count(user_id) as new_number from (select user_id, min(date) as date from login grou...

sql

2021-09-08

0 372

题解 | #牛客每个人最近的登录日期(四)#

sql script select l.date, coalesce(count(user_id),0) from (select distinct date from login) l left join (select user_id, min(date) as date from ...

sql

2021-09-08

0 331

题解 | #牛客每个人最近的登录日期(三)#

# first day login # select user_id, min(date) # from login # group by user_id # second&n...

sql

2021-09-08

0 512

题解 | #牛客每个人最近的登录日期(三)#

sql script # first day login # select user_id, min(date) # from login # group by user_id # second day # select user_id, DATE_ADD(min(date),INTERVA...

sql

2021-09-08

0 305

题解 | #牛客每个人最近的登录日期(二)#

sql script select u.name as u_n, c.name as c_n, tmp.date as date from login l, user u, client c, (select max(date) as date, user_id ...

sql

2021-09-08

0 292

题解 | #牛客每个人最近的登录日期(一)#

sql script select user_id, max(date) as d from login group by user_id order by user_id

sql

2021-09-08

0 290

题解 | #异常的邮件概率#

sql script select e.date, round(sum(case when e.type='no_completed' then 1 else 0 end)/count(e.type),3) from email e where e.send_id in (select id fro...

sql

2021-09-08

0 268

题解 | #找到每个人的任务#

sql script select p.id, p.name, t.content from person p left join task t on p.id=t.person_id

sql

2021-09-07

0 380