Question 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's 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.
I created a query to find the required values and match the expected output to solve this problem.
with user_rank as (select name
from MovieRating m
left join users u
on m.user_id = u.user_id
group by name
order by count(*) desc, name asc
limit 1),
movie_title as (select title
from MovieRating mr
left join Movies mv
on mr.movie_id = mv.movie_id
where month(created_at) = '02' and year(created_at) = 2020
group by title
order by avg(rating) desc, title asc
limit 1)
However, the issue was how to combine these values into a single column. After some research, I found that it is possible using UNION ALL, but I couldn't recall how to use it in this case as the column names were different.
After considering various approaches and conducting some research, I found it to be simpler than I initially thought. It was just a matter of making the column names the same. By renaming the columns as shown below and using UNION ALL, I was able to solve the problem.
select user_rank.name as results
from user_rank
union all
select movie_title.title as results
from movie_title
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