我想要得到的所有结果使用单查询加盟3台

Question

可能重复：
Mysql single query join 3 table and get all the results

嗨，我想列出所有歌曲的专辑，我想列出所有例如，单个歌曲的艺术家见下面。

1. Song Title 1 
    - Artist 1, Artist 2, Artist 3 note:(all this individual artist have link to there artist page) 
2. Song Title 2 
    - Artist 1, Artist 2 
3. Song Title 3 
    - Artist 1, Artist 2, Artist 3 

我的表是歌曲，专辑，艺术家，song_artist

歌曲表

+----+-----------+----------+ 
| id | song_name | album_id | 
+----+-----------+----------+ 
| 1 | Title 1 | 2  | 
| 2 | Title 2 | 2  | 
| 3 | Title 3 | 2  | 
+----+-----------+----------+ 

专辑表

+----+------------+ 
| id | album_name | 
+----+------------+ 
| 1 | Album 1 | 
| 2 | Album 2 | 
| 3 | Album 3 | 
+----+------------+ 

艺术家表

+----+-------------+ 
| id | artist_name | 
+----+-------------+ 
| 1 | Artist 1 | 
| 2 | Artist 2 | 
| 3 | Artist 3 | 
+----+-------------+ 

song_artist表

+--------+--------------+---------+ 
| id  | song_is  |artist_id| 
+--------+--------------+---------+ 
| 1  | 1   | 1  | 
| 2  | 1   | 5  | 
| 3  | 1   | 3  | 
| 4  | 2   | 3  | 
| 5  | 2   | 1  | 
| 6  | 3   | 2  | 
| 7  | 3   | 1  | 
+--------+--------------+---------+ 

这是我当前的代码。

$id =$_GET['id']; *// Get album id from url* 
$query = "SELECT id, song_name, FROM song WHERE album_id = '".$id."'"; 
$result = mysql_query($query) or die("h".mysql_error()); 
while($song = mysql_fetch_assoc($result)){ 
echo $song['song_name']; 
$result1 = mysql_query("SELECT artist.artist_name as artist_name, artist.id as aid 
FROM artist 
INNER JOIN song_artist 
ON artist.id = song_artist.artist_id 
WHERE song_artist.song_id = '".$song['id']."' "); 
while($row = mysql_fetch_array($result1)){ 
echo "<a href='".$row['sid']."'>".$row['artist_name']."</a>, "; 
} 
} 

我该如何编写mysql单个查询来获取php中的所有结果？

非常感谢您的帮助。

James。

好吧现在我得到了我想要的结果，感谢每一个。

$query = "select s.id song_id, s.song_name, group_concat(art.artist_name) artname, a.id 
     from song s 
     left outer join album a on a.id = s.album_id 
     left outer join song_artist sa on sa.song_id = s.id 
     left outer join artist art on art.id = sa.artist_id 
     WHERE a.id= '".$id."' 
     GROUP BY song_id"; 

Answer 1

使用查询：

select s.id song_id, s.song_name, a.album_name, art.artist_name 
from song s 
left outer join album a on a.id=s.album_id 
left outer join song_artist sa on sa.song_id=s.id 
left outer join artist art on art.id=sa.art_id 
where <INSERT WHERE-CLAUSE HERE> 

你会得到这样的结果：

+--------+-----------+------------+-------------+ 
| song_id|song_name | album_name | artist_name | 
+--------+-----------+------------+-------------+ 
| 1  | Title 1 | Album 2 | Artist 1 | 
| 1  | Title 1 | Album 2 | Artist 5 | 
| 1  | Title 1 | Album 2 | Artist 3 | 
| 2  | Title 2 | Album 2 | Artist 3 | 
| 2  | Title 2 | Album 2 | Artist 1 | 
| 3  | Title 3 | Album 2 | Artist 2 | 
| 3  | Title 3 | Album 2 | Artist 1 | 
+--------+-----------+------------+-------------+ 

因此必须迭代每一个ng id从结果中获取每首歌曲的显示信息。因为你只用一个查询就可以做到这一点，所以当一首歌曲有多位艺术家时，你会得到重复的歌曲和专辑名称，但这是你为简洁而付出的代价。

UPDATE：

或者您也可以使用ARTIST_NAME作为GROUP_CONCAT建议Eljakim如果你不会使用艺术家信息任何东西，除了级联显示列表。

Answer 2

这可以让你的歌曲+艺人专辑1.

SELECT song_name, group_concat(artist_name) 
FROM song 
LEFT JOIN song_artist ON song.id=song_artist.song_id 
LEFT JOIN artist ON song_artist.artist_id=artist.id 
WHERE song.album_id=1