题解 | 查询连续登陆的用户

select
        user_id
from (
    select
        user_id,
        count(date_sub_rn) as continue_days
    from (
        select 
            user_id,
            date_sub(date(log_time),interval rn day) date_sub_rn
        from(
            select
                ltb.user_id,
                ltb.log_time,
                row_number() over(partition by user_id order by log_time) as rn
            from 
                    register_tb as rtb
            inner join 
                    login_tb as ltb
            on 
                    rtb.user_id = ltb.user_id
        )a
    )b
    group by user_id
)c
where 
        continue_days > 2
order by 
        user_id