题解 | 查询连续入住多晚的客户信息？

select
    c.user_id,
    g.room_id,
    g.room_type,
    datediff(c.checkout_time, c.checkin_time) as days
from
    checkin_tb as c
    left join guestroom_tb g on c.room_id = g.room_id
where
    datediff(c.checkout_time, c.checkin_time) > 1
order by
    days,
    room_id,
    user_id desc;

datediff:自动忽略时间，只计算日期之间的天数差

where:不能直接使用列的别名。