如何根据条件删除结果以计算平均和特定电影

Posted 2023-02-24

【中文标题】如何根据条件删除结果以计算平均和特定电影

【英文标题】：How do I remove results based on conditions to calculate an average and specific movie

【发布时间】：2021-02-26 14:43:06

【问题描述】：

我有下面的架构。一个简单的解释是：

鲍勃给这部电影评分，5/5 詹姆斯对这部电影的评分提高了 1/5 macy 给这部电影评分，5/5 没有人评价电影复仇者联盟。

逻辑：

如果我是 personA，请查找我已阻止的所有人。 查看所有电影评论。 任何留下了影评且 personA 已屏蔽的人，将其从计算中移除。 计算电影的平均评分。

---

CREATE TABLE movies (
  id integer AUTO_INCREMENT primary key,
  name varchar(100) NOT NULL
);

CREATE TABLE customer (
  id integer AUTO_INCREMENT primary key,
  name varchar(100) NOT NULL
);

CREATE TABLE reviews (
  id integer AUTO_INCREMENT primary key,
  rating integer NOT NULL,
  cus_id integer NOT NULL,
  movie_id integer NOT NULL,
  FOREIGN KEY (cus_id) REFERENCES customer(id),
  FOREIGN KEY (movie_id) REFERENCES movies(id)
);

CREATE TABLE blocked(
  id integer AUTO_INCREMENT primary key,
  cus_id integer NOT NULL, -- This is the person blocking
  blocked_cus_id integer NOT NULL, -- This is the person who is blocked
  FOREIGN KEY (cus_id) REFERENCES customer(id),
  FOREIGN KEY (blocked_cus_id) REFERENCES customer(id)
);

INSERT INTO movies (id, name) VALUES (1, 'up'), (2, 'avengers');
INSERT INTO customer (id, name) VALUES (1, 'bob'), (2, 'james'), (3, 'macy');
INSERT INTO reviews (id, rating, cus_id, movie_id) VALUES (1, 5, 1, 1), (2, 1, 2, 1), (3, 5, 3, 1);
INSERT INTO blocked (id, cus_id, blocked_cus_id) VALUES (1, 1, 2);

我在这里收到了一些关于这个问题的帮助：How do I remove results based on conditions to calculate an average（并且该声明是正确的）但是当我想查找特定电影的评级时，该声明仅显示有评级的电影。我希望它显示这部电影，无论它是否有评级。如果它没有评分，它应该只是说 0。下面，电影复仇者没有评分，也没有显示结果。

SELECT m.name, AVG(r.rating) AS avg_rating
FROM movies m
INNER JOIN reviews r ON m.id = r.movie_id
WHERE NOT EXISTS (SELECT 1 FROM blocked b
                  WHERE b.blocked_cus_id = r.cus_id AND b.cus_id = 1)
AND m.id = 2
GROUP BY m.name;

上面的select语句应该显示：

+----------+------------+
| movie    | avg_rating |
+----------+------------+
| avengers |          0 |
+----------+------------+

当我以 bob 形式查看数据库时，我应该得到：

+-------+------------+
| movie | avg_rating |
+-------+------------+
| up    |          5 |
+-------+------------+

当我将数据库视为 macy 时，我应该得到：

+-------+------------+
| movie | avg_rating |
+-------+------------+
| up    |       3.67 |
+-------+------------+

【参考方案1】：

您想要left join 吗？从您当前的查询开始，这将是：

SELECT m.name, AVG(COALESCE(r.rating, 0)) AS avg_rating
FROM movies m
LEFT JOIN reviews r 
    ON m.id = r.movie_id
    AND NOT EXISTS (
        SELECT 1 
        FROM blocked b
        WHERE b.blocked_cus_id = r.cus_id AND b.cus_id = 1
    )
WHERE m.id = 2
GROUP BY m.id, m.name;