题解 | #查询连续登陆的用户#

select user_id
    from(
    select user_id,date(date_ass),count(user_id) ct
    from(
        select user_id,log_time,date_sub(log_time,interval rk day) date_ass
        from(
            select t1.user_id,log_time,row_number() over(partition by t1.user_id order by log_time) rk
            from register_tb t1 join login_tb t2 on t1.user_id=t2.user_id)
            t)
            a
    group by user_id,date(date_ass))
    b
where ct>=3;