题解 | 查询连续入住多晚的客户信息？

select 
    c.user_id as user_id,
    c.room_id as room_id ,
    g.room_type as room_type,
    datediff(date_format(c.checkout_time,'%Y-%m-%d'), date_format(c.checkin_time,'%Y-%m-%d')) as days
from guestroom_tb g 
    join checkin_tb c 
    on g.room_id = c.room_id
where datediff(date_format(c.checkout_time,'%Y-%m-%d'),date_format(c.checkin_time,'%Y-%m-%d')) > 1
order by days asc, room_id asc, user_id desc