MySQL基础练习题16-电影评分

news2024/12/24 20:19:31

题目

准备数据

分析数据

总结

题目

查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名。
查找在 February 2020 平均评分最高 的电影名称。如果出现平局，返回字典序较小的电影名称。

准备数据

## 创建库
create database db;
use sb;

## 创建movies表
Create table If Not Exists Movies (movie_id int, title varchar(30));

## 创建users表
Create table If Not Exists Users (user_id int, name varchar(30));

## 创建MovieRating表
Create table If Not Exists MovieRating (movie_id int, user_id int, rating int, created_at date);

## 向表中插入数据
Truncate table Movies;
insert into Movies (movie_id, title) values ('1', 'Avengers');
insert into Movies (movie_id, title) values ('2', 'Frozen 2');
insert into Movies (movie_id, title) values ('3', 'Joker');
Truncate table Users;
insert into Users (user_id, name) values ('1', 'Daniel');
insert into Users (user_id, name) values ('2', 'Monica');
insert into Users (user_id, name) values ('3', 'Maria');
insert into Users (user_id, name) values ('4', 'James');
Truncate table MovieRating;
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '1', '3', '2020-01-12');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '2', '4', '2020-02-11');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '3', '2', '2020-02-12');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '4', '1', '2020-01-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '1', '5', '2020-02-17');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '2', '2', '2020-02-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '3', '2', '2020-03-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('3', '1', '3', '2020-02-22');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('3', '2', '4', '2020-02-25');

输入表：

movies表

users表

movierating表

分析数据

结果分两个结果，且不相关，因此最后使用union all进行连接。
1.Daniel 和 Monica 都点评了 3 部电影（"Avengers", "Frozen 2" 和 "Joker"） 但是 Daniel 字典序比较小。

2.Frozen 2 和 Joker 在 2 月的评分都是 3.5，但是 Frozen 2 的字典序比较小。
字典序 ，即按字母在字典中出现顺序对字符串排序，字典序较小则意味着排序靠前。

第一步：查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名。

with t1 as (
    select user_id,count(rating) rating from movierating group by user_id
) select min(a.name) results
  from users a
      join t1 b on b.user_id = a.user_id
where b.rating = (select max(rating) rating from t1);

第二步：查找在 February 2020 平均评分最高 的电影名称。如果出现平局，返回字典序较小的电影名称。

with t2 as (
     select movie_id,avg(rating) rating
 from movierating
 where created_at between '2020-02-01' and '2020-02-29'
 group by movie_id
)select min(a.title) results
 from movies a join t2 b on b.movie_id = a.movie_id
where b.rating = (select max(rating) rating from t2);

第三步：将两个结果进行连接。

with t1 as (
    select user_id,count(rating) rating from movierating group by user_id
) ,t2 as (
     select movie_id,avg(rating) rating
 from movierating
 where created_at between '2020-02-01' and '2020-02-29'
 group by movie_id
)
select min(a.name) results
  from users a
      join t1 b on b.user_id = a.user_id
where b.rating = (select max(rating) rating from t1)
union all
select min(a.title) results
 from movies a join t2 b on b.movie_id = a.movie_id
where b.rating = (select max(rating) rating from t2);