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题解 | #异常的邮件概率#

select e.date,round(sum(case when e.type = 'no_completed' then 1 else 0 end)/count(e.type) ,3) pfrom email e...

2021-09-03

1 329

题解 | #统计salary的累计和running_total#

sum()over(order by ) select emp_no,salary, sum(salary) over(order by emp_no) as running_totalfrom salarieswhere to_date = '9999-01-01';

2021-09-03

1 294

题解 | #获取有奖金的员工相关信息。#

case when + inner join select e.emp_no,e.first_name,e.last_name,eb.btype,s.salary,case when btype = 1 then 0.1salary when btype = 2 then 0.2salary...

2021-09-03

1 336

题解 | #获取所有非manager员工当前的薪水情况#

select de.dept_no,e.emp_no,s.salaryfrom employees ejoin dept_emp deon e.emp_no = de.emp_nojoin salaries son e.emp_no = s.emp_nowhere e.emp_no not in(s...

2021-08-30

1 313

题解 | #获取当前薪水第二多的员工的emp_no以及其对应的薪水salary#

select s.emp_no,s.salary,e.last_name,e.first_namefrom salaries s join employees eon s.emp_no = e.emp_nowhere s.to_date = '9999-01-01'and s.salary in (...

2021-08-29

0 0

题解 | #获得积分最多的人(一)#

select name,sum(grade_num) over(partition by user_id) grade_sumfrom user join grade_info gion user.id = gi.user_idorder by grade_sum desclimit 1;

2021-08-28

0 308

题解 | #实习广场投递简历分析(二)#

select job,left(date,7) as mon, sum(num) as cntfrom resume_infowhere left(date,4) = '2025'group by job,monorder by mon desc,cnt desc;

2021-08-28

1 486

题解 | #牛客的课程订单分析(二)#

select o2.id,o2.user_id,o2.product_name,o2.status,o2.client_id,o2.date from (select *, count(*)over(partition by user_id) as num f...

2021-08-28

0 295

题解 | #牛客的课程订单分析(二)#

select user_idfrom order_infowhere date > '2025-10-15'and status = 'completed'and product_name in('C++','Python','Java')group by user_idhaving coun...

2021-08-26

1 344

题解 | #考试分数(二)#

select g1.id, g1.job, g1.scorefrom grade g1 join ( select job,round(avg(score),3) avg from grade group by job) g2 on g1.job = g2.jobwhere ...

2021-08-26

1 361