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题解 | #查找在职员工自入职以来的薪水涨幅情况#

1.找到最后的薪资 2.找到入职时的工资 3.求差 select q1.emp_no, q1.salary - min as growth from (select emp_no, salary from salaries where to_date = '9999-01-01') q1 join ...

Mysql

2022-04-10

0 272

题解 | #牛客每个人最近的登录日期(五)#

select m.date, round((case when(n.l is not null) then n.l/m.z else 0 end ) ,3) as p from (select t.date, sum(case ranking when 1 then 1 else 0 end...

Mysql

2021-10-09

0 360

题解 | #获取有奖金的员工相关信息。#

select e.emp_no, e.first_name, e.last_name, eb.btype, s.salary, case when(eb.btype = 1) then s.salary * 0.1 when(eb.btype = 2) then s.salary * 0.2 els...

Mysql

2021-10-09

0 333

题解 | #获得积分最多的人(二)#

select t.id, t.name, t.grade_num from( select u.id, u.name, a.grade_num, DENSE_RANK() over (order by a.grade_num desc) as ranking from user u join ( s...

Mysql

2021-10-06

2 418

题解 | #获得积分最多的人(三)#

我人傻了 select t.id, t.name, t.grade_sum from( select total.user_id as id, u.name, total.grade_sum, dense_rank() over (order by grade_sum desc) as rankin...

Mysql

2021-10-06

0 396