题解 | 查询连续入住多晚的客户信息？

select user_id,
       c.room_id,
       room_type,
       datediff(checkout_time,checkin_time) as days
from checkin_tb c join guestroom_tb g on c.room_id = g.room_id
where DATE(checkin_time) >= '2022-06-12' and datediff(checkout_time,checkin_time)>1
order by days,c.room_id,user_id desc

学到了DATE( )函数，也可以用checkin_time >= '2022-06-12 00:00:00'