题解 | 查询连续入住多晚的客户信息？

SELECT
    user_id,
    ct.room_id,
    room_type,
    DATEDIFF (checkout_time, checkin_time) AS days
FROM
    checkin_tb AS ct
    JOIN guestroom_tb AS gt ON ct.room_id = gt.room_id
WHERE
    checkin_time > '2022-06-12'
    AND DATEDIFF (checkout_time, checkin_time) > 1
ORDER BY
    days ASC,
    ct.room_id ASC,
    user_id DESC