题解 | 查询连续入住多晚的客户信息？

select
    A.user_id,
    A.room_id,
    B.room_type,
    (DATE(A.checkout_time)-DATE(A.checkin_time)) AS days
from
    checkin_tb as A
    left join guestroom_tb as B ON A.room_id = B.room_id
where DATE(A.checkin_time) = '2022-06-12'
and (DATE(A.checkout_time)-DATE(A.checkin_time)) > 1
order by (DATE(A.checkout_time)-DATE(A.checkin_time)) asc ,A.room_id asc ,A.user_id desc