这是我的SQL Fiddle
正如您在此处所见，如果我使用DISTINCT，则有2个问题
1.）只有第一个recommendations_vote_average coloum的数字是正确的。所有其他数字的顺序错误
2。）只打印2个号码。
如果我不使用DISTINCT，所有数字都是7.5（即first vote_average）
如何以正确的顺序显示所有（10）数字？
预期输出
movie_title recommendations_vote_average recommendations_title
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
The Dark Knight 7.5,7.5,7.3,8.1,8,7.9,7.9,8,6.6,6.6
SQL小提琴代码：
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
答案 0 :(得分：2)
很难从你的问题中猜出你想要什么。你没有定义它就提到了“正确的顺序”。
You can use GROUP_CONCAT() in these ways.
GROUP_CONCAT(a.b) -- gets all the items in column b -- cardinality preserved
GROUP_CONCAT(DISTINCT a.b) -- distinct values in column b -- cardinality reduced
GROUP_CONCAT(a.b ORDER BY a.b) -- all items in b in order
GROUP_CONCAT(DISTINCT a.b ORDER BY a.b) -- distinct items in b in order
GROUP_CONCAT(a.b ORDER BY a.c) -- all items in b in the same order as c
我不完全确定将DISTINCT添加到最后一个应用程序的含义。
如果您尝试按相应的顺序获取两个连接列，则不能在任何一个中使用DISTINCT; DISTINCT有可能删除重复值。
您的结果集列提到了_average。你得到AVG(value)的实际平均值（算术平均值）。并且它给出了一个单一的聚合数字。
如果您想要一列中的分数列表和另一列中相应的标题列表，请尝试此操作。
GROUP_CONCAT(
recommendations.recommendations_vote_average
ORDER BY recommendations.recommendations_title
) AS recommendations_vote_average,
GROUP_CONCAT(
recommendations.recommendations_title
ORDER BY recommendations.recommendations_title
) AS recommendations_title
按标题顺序显示两个连接列表。
您可能没有意识到这一点：DBMS表中的行没有固有顺序。如果您不止一次说SELECT * FROM table（没有ORDER BY子句），并且每次都以相同的顺序排列，这是一个意外。您的内容中没有任何内容推荐表 - 例如没有唯一的id值 - 给出除分数和标题之外的那些项的顺序。所以你可能无法得到你想要的确切订单。
许多表都包含一个自动增量id列（但不是你的）。在id子句中使用这样的ORDER BY列是一种获得可重复排序的方法。
专业提示：非规范化数据（例如，列中以逗号分隔的数据）通常被认为是有害的。 GROUP_CONCAT()将标准化数据（如输入）转换为非规范化数据。因此，只有在需要时才使用它。
答案 1 :(得分：1)
要获得请求的结果，我认为您需要做的就是删除DISTINCT，但我还建议您引入ORDER BY
SELECT
tmdb_movies.movie_title
, GROUP_CONCAT(r.recommendations_vote_average
ORDER BY r.recommendations_vote_average DESC
SEPARATOR ', '
) as recommendations
, GROUP_CONCAT(r.recommendations_title
ORDER BY r.recommendations_vote_average DESC
SEPARATOR ', '
) as recommendations
FROM tmdb_movies
LEFT JOIN recommendations r ON r.recommendations_tmdb_id=tmdb_movies.tmdb_id
Where tmdb_movies.tmdb_id=1
GROUP BY tmdb_movies.movie_title
;
recommendations | recommendations |
+----+-----------------+----------------------------------------------+---------------------------------------------------------------------------------+
| 1 | The Dark Knight | 8.1, 8, 8, 7.9, 7.9, 7.5, 7.5, 7.3, 6.6, 6.6 | The Lord of the Rings: The Return of the King, Inception, |
| | | | The Lord of the Rings: The The Fellowship of the Ring, The Matrix, |
| | | | The Lord of the Rings: The Two Towers, Batman Begins, |
| | | | The Dark Knight Rises, Iron Man, Captain America: The First Avenger, Iron Man 2 |
+----+-----------------+----------------------------------------------+---------------------------------------------------------------------------------+
如果由我决定，我会将推荐得分与标题相结合（为演示添加手动换行符）：
+----+-----------------+-------------------------------------------------------------------------
| | movie_title | recommendations |
+----+-----------------+-------------------------------------------------------------------------
| 1 | The Dark Knight | 8.1(The Lord of the Rings: The Return of the King);
| 8(Inception); 8(The Lord of the Rings: The The Fellowship of the Ring);
| 7.9(The Matrix); 7.9(The Lord of the Rings: The Two Towers);
| 7.5(Batman Begins); 7.5(The Dark Knight Rises);
| 7.3(Iron Man); 6.6(Captain America: The First Avenger); 6.6(Iron Man 2)
由此查询生成：
SELECT
tmdb_movies.movie_title
, GROUP_CONCAT(DISTINCT concat(r.recommendations_vote_average,'(',r.recommendations_title,')')
ORDER BY r.recommendations_vote_average DESC
SEPARATOR '; '
) as recommendations
FROM tmdb_movies
LEFT JOIN recommendations r ON r.recommendations_tmdb_id=tmdb_movies.tmdb_id
Where tmdb_movies.tmdb_id=1
GROUP BY tmdb_movies.movie_title
更多信息。 以下查询会反转表优先级，并且不使用group_concat
select
m.movie_title, r.*
from recommendations r
left join tmdb_movies m ON r.recommendations_tmdb_id=m.tmdb_id
;
结果：
+----+-----------------+-------------------------+-------------------------------------------------------+------------------------------+
| | movie_title | recommendations_tmdb_id | recommendations_title | recommendations_vote_average |
+----+-----------------+-------------------------+-------------------------------------------------------+------------------------------+
| 1 | The Dark Knight | 1 | The Dark Knight Rises | 7.5 |
| 2 | The Dark Knight | 1 | Batman Begins | 7.5 |
| 3 | The Dark Knight | 1 | Iron Man | 7.3 |
| 4 | The Dark Knight | 1 | The Lord of the Rings: The Return of the King | 8.1 |
| 5 | The Dark Knight | 1 | The Lord of the Rings: The The Fellowship of the Ring | 8 |
| 6 | The Dark Knight | 1 | The Lord of the Rings: The Two Towers | 7.9 |
| 7 | The Dark Knight | 1 | The Matrix | 7.9 |
| 8 | The Dark Knight | 1 | Inception | 8 |
| 9 | The Dark Knight | 1 | Iron Man 2 | 6.6 |
| 10 | The Dark Knight | 1 | Captain America: The First Avenger | 6.6 |
+----+-----------------+-------------------------+-------------------------------------------------------+------------------------------+
样本数据（应该有问题）：
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');