我想知道在 Doctrine2 中处理多对多关系的最好、最干净和最简单的方法是什么。
假设我们有一张像Metallica的Master of Puppets这样的专辑，里面有几首曲目。但请注意，一首曲目可能会出现在一张专辑中，就像Metallica 的Battery一样 - 三张专辑都包含这首曲目。
所以我需要的是专辑和曲目之间的多对多关系，使用带有一些附加列的第三个表（如指定专辑中曲目的位置）。实际上，正如 Doctrine 的文档所建议的那样，我必须使用双重一对多关系来实现该功能。
/** @Entity() */
class Album {
/** @Id @Column(type="integer") */
protected $id;
/** @Column() */
protected $title;
/** @OneToMany(targetEntity="AlbumTrackReference", mappedBy="album") */
protected $tracklist;
public function __construct() {
$this->tracklist = new \Doctrine\Common\Collections\ArrayCollection();
}
public function getTitle() {
return $this->title;
}
public function getTracklist() {
return $this->tracklist->toArray();
}
}
/** @Entity() */
class Track {
/** @Id @Column(type="integer") */
protected $id;
/** @Column() */
protected $title;
/** @Column(type="time") */
protected $duration;
/** @OneToMany(targetEntity="AlbumTrackReference", mappedBy="track") */
protected $albumsFeaturingThisTrack; // btw: any idea how to name this relation? :)
public function getTitle() {
return $this->title;
}
public function getDuration() {
return $this->duration;
}
}
/** @Entity() */
class AlbumTrackReference {
/** @Id @Column(type="integer") */
protected $id;
/** @ManyToOne(targetEntity="Album", inversedBy="tracklist") */
protected $album;
/** @ManyToOne(targetEntity="Track", inversedBy="albumsFeaturingThisTrack") */
protected $track;
/** @Column(type="integer") */
protected $position;
/** @Column(type="boolean") */
protected $isPromoted;
public function getPosition() {
return $this->position;
}
public function isPromoted() {
return $this->isPromoted;
}
public function getAlbum() {
return $this->album;
}
public function getTrack() {
return $this->track;
}
}
样本数据：
Album
+----+--------------------------+
| id | title |
+----+--------------------------+
| 1 | Master of Puppets |
| 2 | The Metallica Collection |
+----+--------------------------+
Track
+----+----------------------+----------+
| id | title | duration |
+----+----------------------+----------+
| 1 | Battery | 00:05:13 |
| 2 | Nothing Else Matters | 00:06:29 |
| 3 | Damage Inc. | 00:05:33 |
+----+----------------------+----------+
AlbumTrackReference
+----+----------+----------+----------+------------+
| id | album_id | track_id | position | isPromoted |
+----+----------+----------+----------+------------+
| 1 | 1 | 2 | 2 | 1 |
| 2 | 1 | 3 | 1 | 0 |
| 3 | 1 | 1 | 3 | 0 |
| 4 | 2 | 2 | 1 | 0 |
+----+----------+----------+----------+------------+
现在我可以显示与它们相关的专辑和曲目列表：
$dql = '
SELECT a, tl, t
FROM Entity\Album a
JOIN a.tracklist tl
JOIN tl.track t
ORDER BY tl.position ASC
';
$albums = $em->createQuery($dql)->getResult();
foreach ($albums as $album) {
echo $album->getTitle() . PHP_EOL;
foreach ($album->getTracklist() as $track) {
echo sprintf("\t#%d - %-20s (%s) %s\n",
$track->getPosition(),
$track->getTrack()->getTitle(),
$track->getTrack()->getDuration()->format('H:i:s'),
$track->isPromoted() ? ' - PROMOTED!' : ''
);
}
}
结果是我所期待的，即：一张专辑列表，其中的曲目按适当的顺序排列，推广的专辑被标记为推广。
The Metallica Collection
#1 - Nothing Else Matters (00:06:29)
Master of Puppets
#1 - Damage Inc. (00:05:33)
#2 - Nothing Else Matters (00:06:29) - PROMOTED!
#3 - Battery (00:05:13)
那么有什么问题呢？
此代码演示了问题所在：
foreach ($album->getTracklist() as $track) {
echo $track->getTrack()->getTitle();
}
Album::getTracklist()返回一个对象数组AlbumTrackReference而不是Track对象。我无法创建代理方法，如果两者兼而有之怎么办，Album并且Track会有getTitle()方法？我可以在方法中做一些额外的处理，Album::getTracklist()但最简单的方法是什么？我是被迫写这样的东西吗？
public function getTracklist() {
$tracklist = array();
foreach ($this->tracklist as $key => $trackReference) {
$tracklist[$key] = $trackReference->getTrack();
$tracklist[$key]->setPosition($trackReference->getPosition());
$tracklist[$key]->setPromoted($trackReference->isPromoted());
}
return $tracklist;
}
// And some extra getters/setters in Track class
编辑
@beberlei 建议使用代理方法：
class AlbumTrackReference {
public function getTitle() {
return $this->getTrack()->getTitle()
}
}
这将是一个好主意，但我正在使用双方的“参考对象”：$album->getTracklist()[12]->getTitle()和$track->getAlbums()[1]->getTitle()，所以getTitle()方法应该根据调用的上下文返回不同的数据。
我将不得不做类似的事情：
getTracklist() {
foreach ($this->tracklist as $trackRef) { $trackRef->setContext($this); }
}
// ....
getAlbums() {
foreach ($this->tracklist as $trackRef) { $trackRef->setContext($this); }
}
// ...
AlbumTrackRef::getTitle() {
return $this->{$this->context}->getTitle();
}
这不是一个非常干净的方式。