牛客9421352号

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题解 | #牛客每个人最近的登录日期(二)#

select u.name as u_n, c.name as c_n, date from login t left outer join user u on t.user_id=u.id left outer join client c on t.client_id=c.id where...

2023-02-09

0 194

题解 | #牛客每个人最近的登录日期(一)#

select user_id, max(date) from login group by user_id order by user_id

2023-02-09

0 176

题解 | #异常的邮件概率#

select date, round(((sum(case when type='completed' then 0 else 1 end)*1.0)/(count(type)*1.0)),3) p from ( select a.*, b.is_blacklist is1, c.is_blac...

2023-02-09

0 216

题解 | #找到每个人的任务#

select a.*, b.content from person a left outer join task b on a.id=b.person_id order by a.id

2023-02-09

0 180

题解 | #刷题通过的题目排名#

#这里的开窗函数自己排了序 select *, dense_rank() over(order by number desc) t_rank from passing_number

2023-02-09

0 197

题解 | #出现三次以上相同积分的情况#

select number from grade group by number having count(number)>=3 order by number

2023-02-09

3 249

给出employees表中排名为奇数行的first_name

#方法一：失败：顺序不对，开窗函数排序对结果产生了影响 #select first_name #from #( #select first_name, #row_number() over(order by first_name) num #from employees #) a #where mo...

2023-02-09

0 174

题解 #统计salary的累计和running_total#

select emp_no, salary, sum(salary) over(rows between unbounded preceding and current row) running_total from salaries where to_date='9999-01-01'

2023-02-09

0 186

题解 | #获取有奖金的员工相关信息。#

select c.*, d.salary, case when c.btype=1 then round(d.salary*0.1,1) when c.btype=2 then round(d.salary*0.2,1) else round(d.salary*0.3,1) ...

2023-02-09

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题解 | #查找在职员工自入职以来的薪水涨幅情况#

#方法一：失败了，可能是由于没有考虑到降薪的情况 #select #emp_no, #maxs-mins growth #from #( # select emp_no, #max(salary) over(partition by emp_no) maxs, #min(salary) over(...

2023-02-09

0 228