题解 | #获取每个部门中当前员工薪水最高的相关信息#

select a.date,(case when c.next_u is null then 0.000
               else round(c.next_u/b.new_u,3) end) p
from (select distinct date from login) a
      left join
     (select date,count(*) new_u
      from login 
      where (user_id,date) in (select user_id,min(date) date from login
                               group by user_id)
      group by date) b
      on a.date = b.date
      left join
     (select date,count(*) next_u
      from login 
      where (user_id,date) in (select user_id,
                               date_add(min(date),interval 1 day) 
                               from login group by user_id)
      group by date) c
      on date_add(b.date,interval 1 day) = c.date
order by a.date;