题解 | #查询连续入住多晚的客户信息？#

select user_id, t1.room_id, room_type, days
from
    (
        select
            user_id,
            room_id,
            DATEDIFF (max(checkout_time), min(checkin_time)) as days
        from
            checkin_tb
        where
            checkin_time >= '2022-06-12 00:00:00'
        group by
            user_id,
            room_id
        having days > 1
    ) t1
    left join guestroom_tb
    on t1.room_id = guestroom_tb.room_id
    order by days, room_id asc, user_id desc;