题解 | 查询连续入住多晚的客户信息？

select t1.user_id, t1.room_id, t2.room_type,
    DATEDIFF(DATE(t1.checkout_time), DATE(t1.checkin_time)) AS days
from checkin_tb t1
join guestroom_tb t2
on t1.room_id = t2.room_id
where substr(t1.checkin_time, 1, 10) >= '2022-06-12'
    and DATEDIFF(DATE(t1.checkout_time), DATE(t1.checkin_time)) > 1
order by days asc, t1.room_id asc, t1.user_id desc