不想上网课的高级磨洋工练习时长两年半

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题解 | 实习广场投递简历分析(一)

select job,sum(num) as cnt from resume_info where year(date) = 2025 group by job order by cnt DESC

2025-10-18

0 17

题解 | 牛客的课程订单分析(三)

select id,user_id,product_name,status,client_id,date from order_info where user_id in (select user_id from order_info where date > '2025-10-15' an...

2025-10-18

0 15

题解 | 牛客的课程订单分析(二)

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

2025-10-18

0 11

题解 | 牛客的课程订单分析(一)

select id,user_id,product_name,status,client_id,date from order_info where year(date) >= 2025 and month(date) >= 10 and day(date)>15 and prod...

2025-10-18

0 14

题解 | 考试分数(四)

select job,floor((max(rk)+1)/2) as start, ceil((max(rk)+1)/2) as end from (select job,score, row_number() over(partition by job order by score DESC)...

2025-10-18

0 14

题解 | 考试分数(三)

select id,name,score from (select id,name,score, dense_rank() over(partition by name order by score DESC) as rk from (select A.id,B.name,A.score from...

2025-10-18

0 17

题解 | 牛客每个人最近的登录日期(四)

select A.date, sum(if(A.date = B.dt,1,0)) as new from login A left join (select user_id, min(date) as dt from login group by user_id) as B on A.user_i...

2025-10-15

0 15

题解 | 牛客每个人最近的登录日期(三)

select round(count(distinct D.user_id)/count(distinct C.user_id),3) as p from login C left join (select A.id,A.user_id,A.client_id,A.date from login ...

2025-10-15

0 14

题解 | 牛客每个人最近的登录日期(二)

select F.u_n,F.c_n,F.dt from (select A.id,A.user_id,B.name as u_n, C.name as c_n, A.date as dt from login A left join user B on A.user_id = B.id left ...

2025-10-15

0 20

题解 | 异常的邮件概率

select dt as date,round(sum(if(re_isb = 0 and send_isb = 0 and type ='no_completed',1,0))/sum(if(re_isb = 0 and send_isb = 0,1,0)),3) as p from (sele...

2025-10-11

0 18