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题解 | 考试分数(四)

select distinct job, case when (count(*) over(partition by job))%2=1 then round(((count(*) over(partition by job))-1)/2+1,0) else round(((...

2025-06-26

0 21

题解 | 考试分数(四)

select distinct job, case when (count(*) over(partition by job))%2=1 then round(((count(*) over(partition by job))-1)/2+1,0) else round(((...

2025-06-26

0 34

题解 | 考试分数(四)

select distinct job, case when (count(*) over(partition by job))%2=1 then round(((count(*) over(partition by job))-1)/2+1,0) else round(((...

2025-06-26

0 21

题解 | 考试分数(三)

select t2.id, t2.name, t2.score from (select t1.id, t1.name, t1.score, dense_rank() over(partition by ...

2025-06-26

0 21

题解 | 考试分数(二)

select t.id, t.job, t.score from (select id, job, score, case when score>avg(score) ove...

2025-06-26

0 20

题解 | 牛客每个人最近的登录日期(六)

select t1.name as u_n, t2.date as date, coalesce(t2.num, 0) as ps_num from user t1 inner join (select user_id, date, ...

2025-06-26

0 19

题解 | 牛客每个人最近的登录日期(五)

with newdetail as( select user_id, date, case when row_number() over(partition by user_id order by date asc)...

2025-06-25

0 31

题解 | 牛客每个人最近的登录日期(四)

select t1.date, sum(if_new) as new from (select l.user_id, l.date, case when row_number() over(partition by l...

2025-06-25

0 27

题解 | 牛客每个人最近的登录日期(三)

select round(count(distinct t2.user_id)/count(distinct t1.user_id),3) as p from (select user_id,date from login)t1 left join (select user_id,date from...

2025-06-24

0 35

题解 | 牛客每个人最近的登录日期(二)

with t1 as (select u.name as u_n, c.name as c_n, l.date, l.user_id from login l left join client c on l.client_id=c.id left join...

2025-06-23

0 34