题解 | 查询连续入住多晚的客户信息？

select
    *
from
    (
        select
            user_id,
            gt.room_id,
            room_type,
            datediff (checkout_time, checkin_time) days
        from
            test.checkin_tb ct
            join test.guestroom_tb gt on ct.room_id = gt.room_id
        where
            date_format (checkin_time, '%Y-%m-%d') > 2022 -06 -12
    ) a
where
    days > 1
order by
    days asc,
    room_id asc,
    user_id desc