题解 | 查询连续入住多晚的客户信息？

with
    chekin_tb_20220612 as(
        select *
        from checkin_tb
        where checkin_time >= '2022-06-12'
        and datediff(checkout_time, checkin_time)>=2
    ),
    guestroom_tb_info as(
        select user_id, chekin_tb_20220612.room_id, room_type, datediff(checkout_time, checkin_time) as days
        from chekin_tb_20220612
        left join guestroom_tb
        on chekin_tb_20220612.room_id = guestroom_tb.room_id
    )
select *
from guestroom_tb_info
order by days asc, room_id asc, user_id desc