Question

我有一个歌曲表，每首歌最多可以有3种不同的流派。所以在我的表中，对于每首歌我都有列genre1，genre2和genre3。我正试图在列表中显示所有类型。

这是一个随机的示例集：

genre1    genre2    genre3
metal     jazz
metal     country   pop
oldies    metal
rap
jazz      hip-hop   choir
choir     metal     jazz

我希望在php中完成的列表按字母顺序显示可供选择的可用不同类型。所以应该列出这个：

合唱
国家
街舞
爵士
金属
老歌
流行音乐
说唱

感谢所有帮助。也许我不是最聪明的方式，但我想不出更好的方法。

Answer 1

所以单独的列没有什么区别？如果是这种情况，您可以使用UNION

SELECT genre1 AS g FROM t UNION SELECT genre2 AS g FROM t UNION SELECT genre3 AS g FROM t

如果您有WHERE子句，那么您需要复制3次，或使用中间临时表

祝你好运！

表：

mysql> SELECT genre1, genre2, genre3 FROM music;
+--------+---------+--------+
| genre1 | genre2  | genre3 |
+--------+---------+--------+
| metal  | jazz    |        | 
| metal  | country | pop    | 
| oldies | metal   |        | 
| rap    |         |        | 
| jazz   | hip-hop | choir  | 
| choir  | metal   | jazz   | 
+--------+---------+--------+
6 rows in set (0.00 sec)

分组：

mysql> SELECT genre1 AS g FROM music UNION ALL
          SELECT genre2 AS g FROM music UNION ALL
          SELECT genre3 AS g FROM music
+---------+
| g       |
+---------+
| metal   | 
| metal   | 
| oldies  | 
| rap     | 
| jazz    | 
| choir   | 
| jazz    | 
| country | 
| metal   | 
|         | 
| hip-hop | 
| metal   | 
|         | 
| pop     | 
|         | 
|         | 
| choir   | 
| jazz    | 
+---------+
18 rows in set (0.00 sec)

数：

mysql> SELECT g, COUNT(*) AS c FROM
             (SELECT genre1 AS g FROM music UNION ALL
              SELECT genre2 AS g FROM music UNION ALL
              SELECT genre3 AS g FROM music)
       AS tg GROUP BY g;
+---------+---+
| g       | c |
+---------+---+
|         | 4 | 
| choir   | 2 | 
| country | 1 | 
| hip-hop | 1 | 
| jazz    | 3 | 
| metal   | 4 | 
| oldies  | 1 | 
| pop     | 1 | 
| rap     | 1 | 
+---------+---+
9 rows in set (0.01 sec)

Answer 2

SELECT genre1, genre2, genre3 FROM table

假设这是一个数组数组，那么：

function coalesce_into_array($aggregate, $row) {
    foreach ($row as $genre) {
        $aggregate[] = $genre;
    }

    return $aggregate;
}

$data = array_unique(array_reduce($data, 'coalesce_into_array', array()));
sort($data);

但是，我不会在严肃的申请中推荐这个。数据库设计很糟糕。阅读有关数据库规范化的信息，了解如何改进它。

Answer 3

除非您出于性能原因将denormalized（2）类型放入三列，否则应该有一个单独的表格来关联歌曲和流派：

CREATE TABLE SongGenres (
    song INT NOT NULL REFERENCES Songs (id) ON DELETE CASCADE,
    genre VARCHAR(32) NOT NULL,
    UNIQUE INDEX (song, genre),
    INDEX genres (genre) -- improves performance for getting genre names
) Engine=InnoDB;

这消除了（“Cross Road Blues”可以在“Blues”和“Delta Blues”下提交，但就此而言）和人工限制（A3的乡村酸屋福音）的三种流派的要求每首歌。如果您有一组有限的流派，则可能需要制作流派列enumerated。 SongGenres表简化了所有类型：

SELECT UNIQUE genre FROM SongGenres;

或者，您可以进一步规范化并为流派创建单独的表：

CREATE TABLE Genres (
    id INT PRIMARY KEY AUTO_INCREMENT,
    name VARCHAR(32) NOT NULL,
    UNIQUE INDEX (name)
) Engine=InnoDB;

CREATE TABLE SongGenres (
    song INT NOT NULL REFERENCES Songs (id) ON DELETE CASCADE,
    genre INT NOT NULL REFERENCES Genres (id) ON DELETE RESTRICT,
    UNIQUE INDEX (song, genre)
) Engine=InnoDB;

这简化了所有类型名称的更多（虽然这只是次要优势）：

SELECT name FROM Genres;

Genres表的主要优点是数据正确性：如果有人拼错了类型，则在Genres表中找不到它。潜在的缺点是它将有效类型限制为表中的类型。当然，为SongGenres提供具有INSERT权限的用户帐户是有意义的，因此这种限制并不严重。一旦你开始添加新类型，你就会遇到与没有类型表时相同的问题：拼写错误。不是添加Genres表中没有的新类型，而是查找类似的类型（使用例如Levenshtein distance或SOUNDS LIKE），如果找到任何类型，请询问用户是否要替换这种类型与发现或保留原始类型之一（并将其添加到流派列表）。

以下是第一种情况下的数据（两个表Songs和SongGenres）：

mysql> SELECT * FROM Songs;
+----+---------------------+--------+----
| id | title               | artist | ...
+----+---------------------+--------+----
|  1 | Cross Road Blues    |  ...
|  2 | Peace In the Valley |  ...   
+----+---------------------+--------+----
2 rows in set (0.00 sec)

mysql> SELECT * FROM SongGenres;
+------+-------------+
| song | genre       |
+------+-------------+
|    2 | acid        |
|    1 | blues       |
|    2 | country     |
|    1 | delta blues |
|    2 | gospel      |
|    2 | house       |
|    2 | techno      |
+------+-------------+
7 rows in set (0.00 sec)

mysql> SELECT s.title, sg.genre FROM Songs AS s JOIN SongGenres AS sg ON s.id=sg.song;
+---------------------+-------------+
| title               | genre       |
+---------------------+-------------+
| Cross Road Blues    | blues       |
| Cross Road Blues    | delta blues |
| Peace In the Valley | acid        |
| Peace In the Valley | country     |
| Peace In the Valley | gospel      |
| Peace In the Valley | house       |
| Peace In the Valley | techno      |
+---------------------+-------------+
7 rows in set (0.00 sec)

使用单独的Genres表，Songs中的数据看起来会相同，但在其他表格中我们会有类似的内容：

mysql> SELECT * FROM Genres;
+----+-------------+
| id | name        |
+----+-------------+
|  1 | acid        |
|  2 | blues       |
|  3 | classical   |
|  4 | country     |
|  5 | delta blues |
|  6 | folk        |
|  7 | gospel      |
|  8 | hip-hop     |
|  9 | house       |
...
| 18 | techno      |
+----+-------------+
18 rows in set (0.00 sec)

mysql> SELECT * FROM SongGenres;
+------+-------+
| song | genre |
+------+-------+
|    1 |     2 |
|    1 |     5 |
|    2 |     1 |
|    2 |     4 |
|    2 |     7 |
|    2 |     9 |
|    2 |    18 |
+------+-------+
7 rows in set (0.00 sec)

mysql> SELECT s.title, g.name AS genre
    ->   FROM Songs AS s 
    ->     JOIN SongGenres AS sg ON s.id=sg.song 
    ->     JOIN Genres AS g ON sg.genre=g.id;
+---------------------+-------------+
| title               | genre       |
+---------------------+-------------+
| Cross Road Blues    | blues       |
| Cross Road Blues    | delta blues |
| Peace In the Valley | acid        |
| Peace In the Valley | country     |
| Peace In the Valley | gospel      |
| Peace In the Valley | house       |
| Peace In the Valley | techno      |
+---------------------+-------------+
7 rows in set (0.00 sec)

Answer 4

SELECT DISTINCT类型，流派，流派3 FROM table

Description

也许您需要更好的数据库设计...... songs | genres | [song_id|genre_id]

如何从多个mysql列中选择不同的值并将它们放在一个PHP数组中？

4 个答案: