题解 | 查询连续入住多晚的客户信息？

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