题解 | #查询连续登陆的用户#

select
    user_id
from
    (
        SELECT
            r.user_id,
            DATE (log_time) as date,
            DENSE_RANK() over (
                PARTITION by
                    r.user_id
                order by
                    DATE (log_time) asc
            ) as ranking
        from
            register_tb r
            join login_tb l on r.user_id = l.user_id
    ) a
where
    ranking >= 3
order by
    user_id asc