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(DATE(checkout_time), DATE(checkin_time)) AS days
FROM checkin_tb AS c
LEFT OUTER JOIN guestroom_tb AS g
ON c.room_id = g.room_id
WHERE DATE(c.checkin_time) >='2022-06-12'
AND DATEDIFF(DATE(checkout_time), DATE(checkin_time)) >=2
ORDER BY days, c.room_id, c.user_id DESC;