1341. Movie Rating

Table: Movies
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| title         | varchar |
+---------------+---------+
movie_id is the primary key (column with unique values) for this table.
title is the name of the movie.
 
Table: Users
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
+---------------+---------+
user_id is the primary key (column with unique values) for this table.
 
Table: MovieRating
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| user_id       | int     |
| rating        | int     |
| created_at    | date    |
+---------------+---------+
(movie_id, user_id) is the primary key (column with unique values) for this table.
This table contains the rating of a movie by a user in their review.
created_at is the user review date. 
 
Write a solution to:
Find the name of the user who has rated the greatest number of movies.
In case of a tie, return the lexicographically smaller user name.
Find the movie name with the highest average rating in February 2020.
In case of a tie, return the lexicographically smaller movie name.
The result format is in the following example.


Example 1:
Input: 
Movies table:
+-------------+--------------+
| movie_id    |  title       |
+-------------+--------------+
| 1           | Avengers     |
| 2           | Frozen 2     |
| 3           | Joker        |
+-------------+--------------+
Users table:
+-------------+--------------+
| user_id     |  name        |
+-------------+--------------+
| 1           | Daniel       |
| 2           | Monica       |
| 3           | Maria        |
| 4           | James        |
+-------------+--------------+
MovieRating table:
+-------------+--------------+--------------+-------------+
| 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  | 
+-------------+--------------+--------------+-------------+
Output: 
+--------------+
| results      |
+--------------+
| Daniel       |
| Frozen 2     |
+--------------+
Explanation: 
Daniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker")
but Daniel is smaller lexicographically.
Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.

 

# 풀이 쿼리

(select u.name as results
  from MovieRating mr
  left join Users u
    on mr.user_id = u.user_id
  group by mr.user_id
  order by count(mr.user_id) desc, name
  limit 1)

union all

(select m.title as results
  from MovieRating mr
  left join Movies m
    on mr.movie_id = m.movie_id
  where created_at between '2020-02-01' and '2020-02-29'
  group by m.title
  order by avg(mr.rating) desc, m.title
  limit 1)

 

- 유저 이름과 영화 이름은 각각 구했는데, 이 둘을 어떻게 한번에 출력할 것인가를 좀 고민했다.

이 방법을 써야 하나 저 방법을 써야 하나...

 

Union all이 없었으면 아마 한참 고생했을 것이다.

 

이 문제의 핵심은 유저 이름과 영화 이름을 구하는 조건을 각각 잘 구현해서 출력하고, 이 둘을 union all로 한 번에 출력하는 것이었다.