SQL查詢以正確順序顯示2列

Question

查看我在此提琴中的SQL查詢：sql fiddle

我想從SQL查詢中獲得什麼？

正如你所看到的，我在recommendations表中插入3件東西。

1. recommendations_tmdb_id
2. recommendations_title
3. recommendations_vote_average

我用recommendations_tmdb_id到recommendations表連接到`tmdb_movies表

我想顯示recommendations_title和recommendations_vote_average。

在我原來的代碼：我這是如何顯示的數據：

The Dark Knight Rises - 7.5 
Batman Begins - 7.5 
Iron Man - 7.3 
The Lord of the Rings: The Return of the King - 8.1 

等。

但是爲了簡單起見，我不希望包括完整的代碼，只是所需碼。

所以在SQL小提琴，這是數據如何迴應

問題與此代碼：

問題一：

仔細看看代碼。第二recommendations_vote_average數爲7.5和第二recommendations_title電影的名字是Batman Begins

所以，這意味着我的代碼顯示蝙蝠俠電影中有一個vote_average評級7.5，這是假的。我插入蝙蝠俠開始評分爲7.3而不是7.5。

它顯示一些其他電影的評級。所以這意味着，以錯誤的順序。

問題號2：

它僅示出了6 vote_average結果，而不是10。我插入10個記錄。 爲什麼只顯示6條記錄而不是10條記錄？因爲我的group_concat行中有DISTINCT。

如果我刪除DISTINCT會怎麼樣？然後，它會向所有電影顯示7.5的評分。

那麼，我該如何解決這個問題？

預期結果：

 recommendations_vote_average       recommendations_title 


7.5, 7.5, 7.3, 8.1, 8, 7.9, 7.9, 8, 6.6, 6.6   The Dark Knight Rises,Batman Begins,Iron Man,The Lord of the Rings: The Return of the King,The Lord of the Rings: The The Fellowship of the Ring,The Lord of the Rings: The Two Towers,The Matrix,Inception,Iron Man 2,Captain America: The First Avenger 

SQL小提琴全碼

CREATE TABLE tmdb_movies (
    tmdb_id INTEGER NOT NULL PRIMARY KEY, 
    movie_title TEXT NOT NULL 
); 

INSERT INTO tmdb_movies (tmdb_id, movie_title) VALUES 
(1, 'The Dark Knight'); 


CREATE TABLE recommendations (
    recommendations_tmdb_id INTEGER NOT NULL, 
    recommendations_title TEXT NOT NULL, 
    recommendations_vote_average TEXT NOT NULL 
); 




INSERT INTO recommendations (recommendations_tmdb_id, recommendations_title, recommendations_vote_average) VALUES 
(1, 'The Dark Knight Rises', '7.5'), 
(1, 'Batman Begins', '7.5'), 
(1, 'Iron Man', '7.3'), 
(1, 'The Lord of the Rings: The Return of the King', '8.1'), 
(1, 'The Lord of the Rings: The The Fellowship of the Ring', '8'), 
(1, 'The Lord of the Rings: The Two Towers', '7.9'), 
(1, 'The Matrix', '7.9'), 
(1, 'Inception', '8'), 
(1, 'Iron Man 2', '6.6'), 
(1, 'Captain America: The First Avenger', '6.6'); 


SELECT tmdb_movies.movie_title 
,GROUP_CONCAT(DISTINCT recommendations.recommendations_vote_average) as recommendations_vote_average 
,GROUP_CONCAT(DISTINCT recommendations.recommendations_title) as recommendations_title 
FROM tmdb_movies 

LEFT JOIN recommendations ON recommendations.recommendations_tmdb_id=tmdb_movies.tmdb_id 

Where tmdb_movies.tmdb_id=1 

GROUP BY tmdb_movies.movie_title 

Answer 1

降DSTINCT從GROUP_CONCAT（）

CREATE TABLE tmdb_movies (
    tmdb_id INTEGER NOT NULL PRIMARY KEY, 
    movie_title TEXT NOT NULL 
); 

INSERT INTO tmdb_movies (tmdb_id, movie_title) VALUES 
(1, 'The Dark Knight'); 


CREATE TABLE recommendations (
    recommendations_tmdb_id INTEGER NOT NULL, 
    recommendations_title TEXT NOT NULL, 
    recommendations_vote_average TEXT NOT NULL 
); 




INSERT INTO recommendations (recommendations_tmdb_id, recommendations_title, recommendations_vote_average) VALUES 
(1, 'The Dark Knight Rises', '7.5'), 
(1, 'Batman Begins', '7.5'), 
(1, 'Iron Man', '7.3'), 
(1, 'The Lord of the Rings: The Return of the King', '8.1'), 
(1, 'The Lord of the Rings: The The Fellowship of the Ring', '8'), 
(1, 'The Lord of the Rings: The Two Towers', '7.9'), 
(1, 'The Matrix', '7.9'), 
(1, 'Inception', '8'), 
(1, 'Iron Man 2', '6.6'), 
(1, 'Captain America: The First Avenger', '6.6'); 


SELECT tmdb_movies.movie_title 
,GROUP_CONCAT(recommendations.recommendations_vote_average) as recommendations_vote_average 
,GROUP_CONCAT(recommendations.recommendations_title) as recommendations_title 
FROM tmdb_movies ; 

返回

movie_title recommendations_vote_average recommendations_title 
1 The Dark Knight 7.5,7.5,7.3,8.1,8,7.9,7.9,8,6.6,6.6 The Dark Knight Rises,Batman Begins,Iron Man,The Lord of the Rings: The Return of the King,The Lord of the Rings: The The Fellowship of the Ring,The Lord of the Rings: The Two Towers,The Matrix,Inception,Iron Man 2,Captain America: The First Avenger 

http://rextester.com/NNJA85394