题解 | 查询连续入住多晚的客户信息？

select c.user_id,c.room_id,g.room_type,datediff(checkout_time,checkin_time) days
from checkin_tb c
left join guestroom_tb g on c.room_id=g.room_id
where c.checkin_time>='2022-06-12'
and datediff(checkout_time,checkin_time) >=2
order by days asc,room_id asc,user_id desc;

1、datediff()

大的时间在前面

2、从6.12起：

where c.checkin_time>='2022-06-12',不用date_format也能识别