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题解 | #牛客的课程订单分析(二)#

select user_id from order_info where status='completed' and product_name in ('C++','Java','Python') and datediff(date,'2025-10-15')>0 group by user...

2024-07-29

0 158

题解 | #牛客的课程订单分析(一)#

select * from order_info where datediff(date,'2025-10-15')>0 and product_name in ('C++','Java','Python') and status='completed' order by id 注意筛选条件...

2024-07-29

0 128

题解 | #考试分数(五)#

select b.id, c.job, b.score, b.t_rank from ( select a.job, ceil(count_ / 2) start, cei...

2024-07-29

0 153

题解 | #考试分数(四)#

select a.job, ceil((a.count_)/2) start, ceil((a.count_+1)/2) end from (select job,count(*) count_ from grade group by job) a order by a.job 求中位数位置利...

2024-07-29

0 116

题解 | #考试分数(三)#

select a.id,a.name,a.score from (select g.id,l.name,g.score,dense_rank() over(partition by g.language_id order by score desc) rank_ from grade g left ...

2024-07-29

0 124

题解 | #考试分数(二)#

select a.id,a.job, a.score from (select id, job, score, avg(score) over(partition by job) avg_score from grade order by id) a where a.score > a.avg...

2024-07-27

1 127

题解 | #考试分数(一)#

select job, round(avg(score),3) avg from grade group by job order by avg desc 考察聚合函数的使用

2024-07-27

1 131

题解 | #牛客每个人最近的登录日期(六)#

select u.name u_n,p.date,sum(p.number) over(partition by u.id order by date) from passing_number p left join user u on p.user_id=u.id order by p.date,...

2024-07-27

1 144

题解 | #牛客每个人最近的登录日期(五)#

select a.date, ifnull(round(count(l.user_id)/count(b.user_id),3),0.0) from (select distinct date from login ) a left join (select user_id,min(date) da...

2024-07-27

1 127

题解 | #牛客每个人最近的登录日期(四)#

select a.date, count(b.user_id) from (select distinct date from login) a left join (select user_id,min(date) date from login group by user_id) b on a....

2024-07-27

1 97