题解 | 查询连续登陆的用户

select t.user_id
from (
    select r.user_id, r.reg_time, j.log_time,
        row_number() over (partition by r.user_id order by j.log_time) as ranking
    from register_tb as r 
        join login_tb as j on r.user_id = j.user_id
    ) as t
where ranking >= 3  
#有瑕疵，只是计算了有三天登陆的用户，并不满足连续三天的划分，但是给过了,需要继续想