题解 | 查询连续登陆的用户

我这个想法挺巧妙的：把用户登录天数排序，然后用天数-序数，这样就得到了一个数字。
最后用group by 查看每个人每个数字重复次数，就筛选出了连续登录天数
select user_id
from
(select user_id,ranking,count(1) as cnt
from 
(select user_id,day(log_time)-
row_number()over(partition by user_id order by log_time) as ranking
from
(    select a.user_id,log_time,log_port
    from login_tb a ,register_tb b
    where a.user_id=b.user_id
    union all
    select user_id,reg_time as log_time,reg_port as log_port
    from register_tb

    order by user_id,log_time) as t) as t2
group by user_id,ranking
having cnt>=3) as t3