题解 | #查询连续入住多晚的客户信息？#

这题比较容易，就是难在排序了，需要多注意
with tiaojian as (
select 
user_id,
cb.room_id,
room_type,
sum(datediff(checkout_time,checkin_time)) as pdiff
from checkin_tb cb left join guestroom_tb tb 
on cb.room_id=tb.room_id
where
date(checkin_time)>="2022-06-12"
group by room_id,room_type,user_id
)


select 
user_id,
room_id,
room_type,
pdiff as days
from tiaojian
where
pdiff>1
order by days ,room_id,user_id desc