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PHP MySQL将GROUP_CONCAT结果视为单独的项目

[英]PHP MySQL Treat GROUP_CONCAT results as separate items

I am making a music classification database using PHP and MySQL. 我正在使用PHP和MySQL创建音乐分类数据库。 I have three tables: 我有三个表:

Song: 歌曲:

ID  | Title
------------------------
1   | Example Song

Genre: 类型:

ID  | Name
------------------------
1   | Classical
2   | Instrumental

SongGenre 歌曲类型

SongID | GenreID
----------------
1      | 1
1      | 2

My Query is: 我的查询是:

SELECT s.title, GROUP_CONCAT(DISTINCT g.name SEPARATOR ', ')
FROM song s    
LEFT JOIN songgenre sg ON s.id=sg.s_id
LEFT JOIN genre g ON sg.genreid = g.id

I'm using GROUP_CONCAT to allow for multiple genres as shown: 我正在使用GROUP_CONCAT允许多种流派,如下所示:

Title: "Example Song" Genres: Classical, Instrumental 标题:“歌曲范例”类型:古典,器乐

I wish to generate a link in PHP for each genre, so that if the user clicks on "Classical" they are brought to more songs listed as Classical. 我希望为每种流派在PHP中生成一个链接,这样,如果用户单击“古典”,它们就会被带到更多列为古典的歌曲中。 The issue is, I am unsure how to give each genre its' own link. 问题是,我不确定如何赋予每种类型自己的链接。 The issue with GROUP_CONCAT is that both genres are returned together in the same row, and I am unsure how to split the row apart to add a link to each separate genre. GROUP_CONCAT的问题在于两种类型都在同一行中一起返回,我不确定如何将行分开以为每个单独的类型添加链接。

Don't group within the database layer—return an ungrouped ( but sorted ) recordset to PHP and handle it from there: 不要在数据库层内分组-将未分组( 但已排序 )的记录集返回给PHP并从那里进行处理:

$qry = $pdo->query('
  SELECT   sg.SongID, sg.GenreID, s.Title, g.Name
  FROM     song s
             LEFT JOIN songgenre sg ON s.ID = sg.SongID
             LEFT JOIN genre g ON sg.GenreID = g.ID
  ORDER BY sg.SongID, sg.GenreID
');

if ($qry) {
  echo '<ul class="songs">';

  $row = $qry->fetch();
  while ($row) {
    $current_song = $row['SongID'];

    echo '<li>'
       ,   '<span class="title">', htmlentities($row['Title']), '</span>'
       ,   '<ul class="genres">';
    do {
      echo   '<li>'
         ,     '<a href="genre.php?id=', intval($row['GenreID']), '">'
         ,       htmlentities($row['Name'])
         ,     '</a>'
         ,   '</li>';
    } while ($row = $qry->fetch() and $row['SongID'] == $current_song);

    echo   '</ul>'
       , '</li>';
  }

  echo '</ul>';
}

a very basic example as requested by OP OP要求的一个非常基本的示例

$var="Classical, Instrumental";

$each=explode(', ',$var);

foreach($each as $v){
echo '<a href="search.php?genre='.$v.'">'.$v.'</a>';
}

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