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全部文章 / 题解（共24篇）

题解 | #统计作答次数#

select sum(case when start_time is not null then 1 end) ,sum(case when submit_time is not null and score is not null then 1 end) ,count(distinct case ...

Mysql

2021-11-13

0 235

题解 | #SQL类别高难度试卷得分的截断平均值#

select tb1.tag ,tb1.difficulty ,round(avg(tb1.score),1) from( select er.score ,ei.tag ,ei.difficulty ,row_number() over(order by er.score desc) r1 ,ro...

Mysql

2021-11-13

0 302

题解 | #最差是第几名(一)#

selecta.grade,sum(a.number) over (order by grade)from ( select * from class_grade order by grade) a

2021-08-07

0 332

题解 | #实习广场投递简历分析(三)#

selectb.job,b.fm,b.fs,c.sm,c.ssfrom( select a.job ,a.mon fm ,sum(a.num) fs from ( select id ,job ,substring...

2021-08-07

0 357

题解 | #实习广场投递简历分析(二)#

selecta.job,a.mon,sum(a.num) rfrom ( select id ,job ,substring(date,1,7) mon ,num from resume_info where date >= '2025-01-01' ...

2021-08-07

0 356

题解 | #牛客的课程订单分析(六)#

selecta.id,a.is_group_buy,c.namefrom ( select * from order_info where user_id in ( select user_id from order_info ...

2021-08-07

0 355

题解 | #牛客的课程订单分析(五)#

selectb.user_id,b.m,a.date,b.cfrom ( select * ,row_number() over(partition by user_id order by date) r from order_info where date > ...

2021-08-05

0 369

题解 | #考试分数(四)#

selecta.job,case when mod(max(a.r),2) = 0 then ceiling(max(a.r)/2) else ceiling(max(a.r)/2) end,case when mod(max(a.r),2) = 0 then ceiling(max(a.r)/2+...

2021-08-03

0 306

题解 | #考试分数(三)#

selecta.id,a.name,a.scorefrom ( select g.id ,l.name ,g.score ,dense_rank() over(partition by l.name order by g.score desc) r from gr...

2021-08-03

0 298

题解 | #牛客每个人最近的登录日期(六)#

select*from( select n.name ,l.date ,sum(p.number) over(partition by l.user_id order by l.date) s from login l join user n on l.us...

2021-08-03

0 336