“最长的影片租借时间和租借该影片的客户” MySQL Sakila 查询

Question

我正在尝试编写一个查询，该查询将返回 Sakila 数据库中最长的电影租借时间和租借该电影的客户（以及电影的标题）。

作为第一步，我编写了以下查询，它应该返回film_id、inventory_id、customer_id 和最长租期（以周为单位）：

SELECT DDIFF.*

FROM (SELECT rental.inventory_id, rental.customer_id, DATEDIFF(rental.return_date, rental.rental_date)/7 AS TM FROM rental) AS DDIFF

INNER JOIN  (SELECT  DDIFF.inventory_id, DDIFF.customer_id, MAX(DDIFF.TM) AS WEEKS 
FROM (SELECT  rental.inventory_id, rental.customer_id, DATEDIFF(rental.return_date, rental.rental_date)/7 AS TM FROM rental) AS DDIFF
GROUP BY DDIFF.inventory_id, DDIFF.customer_id) AS MXLST 

ON  DDIFF.TM = MXLST.WEEKS AND  MXLST.inventory_id = DDIFF.inventory_id AND MXLST.customer_id = DDIFF.customer_id

;

但我得到了一些意想不到的结果（它应该返回 multiple 行，最长持续时间为 1.4286 周，但由于某种原因我得到了 1.0 周）。我究竟做错了什么？如果能听到解释，我将不胜感激。

请注意，我已尝试按照 this 和 this 正确使用 MAX() 但它仍然没有返回正确的结果。

编辑：Sakila 表的结构可以在here找到

Answer 1

似乎我已经设法解决了我自己的问题。这是一个更新的查询以供将来参考：

SELECT customer.first_name,
       customer.last_name,
       film.title,
       DDIFF.TM AS 'rental duration'
FROM customer,
     film,
     (SELECT inventory.film_id,
             rental.inventory_id,
             rental.customer_id,
             DATEDIFF(rental.return_date, rental.rental_date)/7 AS TM
      FROM rental,
           inventory
      WHERE rental.inventory_id = inventory.inventory_id) AS DDIFF

INNER JOIN (SELECT DDIFF.film_id,
              DDIFF.inventory_id,
              DDIFF.customer_id,
              MAX(DDIFF.TM) AS WEEKS 
            FROM (SELECT inventory.film_id,
                         rental.inventory_id,
                         rental.customer_id,
                         DATEDIFF(rental.return_date, rental.rental_date)/7 AS TM
                  FROM rental,
                       inventory
                  WHERE rental.inventory_id = inventory.inventory_id) AS DDIFF
            GROUP BY DDIFF.film_id,
                     DDIFF.inventory_id,
                     DDIFF.customer_id) AS MXLST 
ON  DDIFF.TM = (SELECT MAX(DDIFF.TM)
                FROM (SELECT inventory.film_id,
                             rental.inventory_id,
                             rental.customer_id,
                             DATEDIFF(rental.return_date, rental.rental_date)/7 AS TM
                      FROM rental,
                           inventory
                      WHERE rental.inventory_id = inventory.inventory_id) AS DDIFF)
AND MXLST.film_id = DDIFF.film_id
AND MXLST.inventory_id = DDIFF.inventory_id
AND MXLST.customer_id = DDIFF.customer_id

WHERE customer.customer_id = DDIFF.customer_id
AND film.film_id = DDIFF.film_id;

我承认，它效率不高，看起来也不像样，但这是一个正确的答案。欢迎任何人尝试找到返回相同结果的更有效/可读性更高的查询。无论如何，感谢所有试图提供帮助的人。

Answer 2

您的查询非常低效：您正在对完整表执行 3 SELECTs，然后执行其他操作。我测试了你的查询：它已经运行了 60 多秒而没有给出任何结果。刚刚杀死它并重写您的查询如下：

SELECT c.first_name, c.last_name, f.title, DATEDIFF(r.return_date,r.rental_date)/7 AS rental_time 
FROM rental AS r
INNER JOIN inventory AS i 
ON r.inventory_id = i.inventory_id
INNER JOIN film AS f
ON i.film_id = f.film_id
INNER JOIN customer AS c
ON r.customer_id = c.customer_id
HAVING rental_time = (SELECT MAX(DATEDIFF(return_date,rental_date))/7 FROM rental)