切片一個MySQL陣列

Question

我有一個MySQL查詢

$raw_resultsplaylists = mysql_query("SELECT songs.*, playlists.title as `playlist_title` 
           FROM song_main AS songs 
           JOIN songs_in_list AS playlist_songs 
            ON songs.ID = playlist_songs.song_id 
           JOIN playlists 
           ON playlist_songs.playlist_id = playlists.ID 
           WHERE playlists.owner_id = '$userid'") or die(mysql_error()); 

檢索的信息，如一個數組：

playlist_title  song_name    song_url  song_artist 
    -------------------------------------------------------------------   
    foo1    woops     www.    - 
    foo1    blah     www.    - 
    foo2    blop     www.    - 
    foo1    woop     www.    - 
    foo3    bob     www.    -  
    foo1    dylan     www.    - 

我的問題是我怎麼可以切片Mysql的陣列來獲得所有的行在同一playlist_title在一個單獨的陣列，所以我可以在單獨的html表格中顯示它們，例如：

html表格1：

playlist_title  song_name    song_url  song_artist 
    -------------------------------------------------------------------   
    foo1    woops     www.    - 
    foo1    blah     www.    - 
    foo1    woop     www.    -  
    foo1    dylan     www.    - 

HTML表2

playlist_title  song_name   song_url  song_artist 
    -------------------------------------------------------------------   
    foo2    blop     www.    - 
    foo2    dylan     www.    - 

等等

Answer 1

剛繪製的想法..

do{ 

    $groupedItems[$row['pname']][] = $row; //every row with the same pname will be "grouped" in the same array level 

}while.. 

然後......

foreach($groupedItems as $group){ 
    //bla bla 
} 

Answer 2

有沒有 「陣列」 中的SQL。這是所有的行和列。你想要的是，也許是：

SELECT * FROM table_name WHERE pname='foo1' 

或許：

SELECT * FROM table_name ORDER BY pname 

在那裏你可以做你的應用程序中的分裂。

Answer 3

對於HTML表1：

mysql_query("SELECT songs.*, playlists.title as `playlist_title` 
          FROM song_main AS songs 
          JOIN songs_in_list AS playlist_songs 
          ON songs.ID = playlist_songs.song_id 
          JOIN playlists 
          ON playlist_songs.playlist_id = playlists.ID 
          WHERE playlists.owner_id = '$userid' 
          AND playlist_title='foo1'") or die(mysql_error());