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题解 | #异常邮件的概率#

代码量最多但是最无脑的写法…… 选出所有正常用户发送的邮件 select a.date,round(b.bad/a.num,3) as pfrom(select date,count(id) num from emailwhere send_id in (select id from user wh...

Mysql

2021-09-25

0 333

题解 | #每个人的任务#

select a.id,a.name,b.contentfrom person aleft join task bon a.id = b.person_idorder by a.id 使用left join 保留全部左边的person即可这里需要注意比较坑的是 task表的person_id 对应...

Mysql

2021-09-25

0 404

题解 | #刷题排名#

select id,number,(select count(DISTINCT number)+1 from passing_number p2 where p1.number<p2.number) t_rank from pas...

Mysql

2021-09-25

0 296

题解 | #查找在职员工自入职以来的薪水涨幅情况#

select a.emp_no,(b.salary-a.salary) growth ##同一行中两列相减即为涨幅 from (select emp_no,salary from salaries ###第三层这里选出在职员工入职时的工资 where from_date in (sel...

Mysql

2021-09-17

0 359