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mysql - 查找对电影评分最多的用户的姓名

问题描述

表：电影

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| title         | varchar |
+---------------+---------+

movie_id 是该表的主键。标题是电影的名称。表：用户

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
+---------------+---------+

user_id 是该表的主键。表：Movie_Rating

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| user_id       | int     |
| rating        | int     |
| created_at    | date    |
+---------------+---------+

(movie_id, user_id) 是该表的主键。此表包含用户在其评论中对电影的评分。created_at 是用户的评论日期。

编写以下 SQL 查询：

查找对电影评分最多的用户的姓名。如果出现平局，则返回按字典顺序较小的用户名。

查找 2020 年 2 月平均评分最高的电影名称。如果出现平局，则返回字典顺序较小的电影名称。

查询分2行返回，查询结果格式如下例：

电影表：

+-------------+--------------+
| movie_id    |  title       |
+-------------+--------------+
| 1           | Avengers     |
| 2           | Frozen 2     |
| 3           | Joker        |
+-------------+--------------+

用户表：

+-------------+--------------+
| user_id     |  name        |
+-------------+--------------+
| 1           | Daniel       |
| 2           | Monica       |
| 3           | Maria        |
| 4           | James        |
+-------------+--------------+

Movie_Rating 表：

+-------------+--------------+--------------+-------------+
| movie_id    | user_id      | rating       | created_at  |
+-------------+--------------+--------------+-------------+
| 1           | 1            | 3            | 2020-01-12  |
| 1           | 2            | 4            | 2020-02-11  |
| 1           | 3            | 2            | 2020-02-12  |
| 1           | 4            | 1            | 2020-01-01  |
| 2           | 1            | 5            | 2020-02-17  | 
| 2           | 2            | 2            | 2020-02-01  | 
| 2           | 3            | 2            | 2020-03-01  |
| 3           | 1            | 3            | 2020-02-22  | 
| 3           | 2            | 4            | 2020-02-25  | 
+-------------+--------------+--------------+-------------+

结果表：

+--------------+
| results      |
+--------------+
| Daniel       |
| Frozen 2     |
+--------------+

丹尼尔和莫妮卡已经对 3 部电影（“复仇者联盟”、“冰雪奇缘 2”和“小丑”）进行了评级，但丹尼尔在词典上更小。《冰雪奇缘 2》和《小丑》在 2 月份的平均评分为 3.5，但《冰雪奇缘 2》的字典顺序较小。

答案1：

(select name results
from users
where user_id in (
    select user_id
    from movie_rating
    group by user_id
    having count(user_id) = (
        select max(cnt)
        from (
            select count(user_id) cnt
            from movie_rating
            group by user_id) f1))
order by results
limit 1)
union 
(select title results
from movies
where movie_id in(
    select movie_id
    from movie_rating
    where left(created_at,7) = '2020-02'
    group by movie_id
    having avg(rating) =(
        select max(avgr) 
        from (
            select avg(rating) avgr
            from movie_rating
            where left(created_at,7) = '2020-02'
            group by movie_id)f2))
order by results
limit 1)

答案2：

(
select name
from users u
join (
    select user_id, count(*) cnt
    from movie_rating
    group by user_id
    order by cnt
    limit 1) f1
on u.user_id = f1.user_id)
union
(
select title
from movies m
join (
    select movie_id, avg(rating) avgr
    from movie_rating
    where left(created_at,7) = '2020-02'
    group by movie_id
    order by avgr
    limit 1) f2
on m.movie_id = f2.movie_id)

两者都是错误的

标签： mysqlsql-server

解决方案

这应该这样做。它在 SQLite 中。

SELECT name FROM (
   SELECT u.name name, count(*) total 
   FROM users u, movie_rating mr 
   WHERE u.user_id=mr.user_id
   GROUP BY u.name
) ORDER BY total desc, name 
LIMIT 1
UNION
SELECT title FROM (
  SELECT m.title title, AVG(mr.rating) average 
  FROM movies m,  movie_rating mr 
  WHERE m.movie_id=mr.movie_id 
    AND created_at BETWEEN '2020-02' and '2020-03' 
  GROUP BY title
) ORDER BY average desc, title 
LIMIT  1;

注意策略。用内部查询中的名称构造聚合数，然后在外部查询中排序并选择顶部

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