Question

我正在尝试回答一个具体问题：““运动”类别中哪部电影最受欢迎？” 我已经尝试过

WITH CustomerRentalsPerStore AS
(
    SELECT  R.customer_id, I.category_id,  COUNT (R.inventory_id) as rental_count 
    from    rental AS R
            INNER JOIN inventory AS I 
                on R.inventory_id = I.inventory_id
    GROUP BY customer_id, I.category_id
    --ORDER BY COUNT (R.inventory_id) desc
)

SELECT c.customer_id, c.first_name, c.last_name, cr.rental_count, cr.store_id 
FROM    Customer C
        INNER JOIN CustomerRentalsPerStore CR
            on  C.customer_id = CR.customer_id
where   cr.rental_count = (SELECT MAX(rental_count) FROM CustomerRentalsPerStore)       
 AND CR.category_id='Sports'

以下是ER图：

任何帮助将不胜感激！谢谢

Answer 1

根据您为我澄清的运动类电影中租金最高的电影 ，我有以下 estested SQL应该可以为您提供结果：

SELECT  f.title, COUNT(*) AS RentalCount
  FROM  film f
    INNER JOIN film_category fc ON fc.film_id = f.film_id
    INNER JOIN category c ON c.category_id = fc.category_id
    INNER JOIN inventory i ON i.film_id = f.film_id
    INNER JOIN rental r ON r.inventory_id = i.inventory_id
  WHERE (c.name = 'Sports')
  GROUP BY f.title
  ORDER BY 2 DESC;

这可以有效地获得“体育”类别中所有电影的所有租赁次数（COUNT）。显然，您只希望第一个结果，因此仅将输出限制为一行。

该代码未经测试，但应为您指明正确的方向。

Answer 2

另一种方法是在进入GROUP BY和COUNT(*)处理之前，首先使选择尽可能小/限制。
优化器可能会选择一种更好的执行方法，但这取决于索引。

    SELECT 
       film.title  
     , COUNT(*)
    FROM (
      SELECT 
        category.category_id
      FROM 
        category    
      INNER JOIN
        film_category
      ON
        category.category_id = film_category.category_id   
      INNER JOIN 
       film 
      ON
        film_category.category_id = film.film_id
      INNER JOIN
        inventory 
      ON
        film.film_id = inventory.film_id 
      INNER JOIN
        rental 
      ON
        inventory.film_id= rental.inventory_id   
      WHERE
        category.name = 'Sports'     
    ) AS alias
    INNER JOIN 
     film
    ON
      alias.film_id = film.film_id
    GROUP BY 
     film.title   
    ORDER BY
     COUNT(*) DESC
   LIMIT 1

PostgreSQL-在“运动”类别中哪部电影最受欢迎？

2 个答案: