Laravel 4，来自一个模型的多个多态关系
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Laravel 4, Multiple Polymorphic Relations from one Model
提问by Joe
I'm trying to set up polymorphic relationships in Laravel 4 so that I can have one Image class which handles everything related to uploads, unlinks and so on, then have it used by multiple different Models. That's fine, until I get to trying to create multiple links from the same Model.
我正在尝试在 Laravel 4 中设置多态关系，以便我可以拥有一个 Image 类来处理与上传、取消链接等相关的所有内容，然后让多个不同的模型使用它。没关系，直到我开始尝试从同一个模型创建多个链接。
For example, I currently have something like this:
例如，我目前有这样的事情：
Models/Image.php
模型/Image.php
class Image extends Eloquent {
public function of() { return $this->morphTo(); }
}
Models/Person.php
模型/Person.php
class Person extends Eloquent {
public function mugshot() { return $this->morphOne('Image', 'of'); }
public function photos() { return $this->morphMany('Image', 'of'); }
}
Models/Place.php
模型/Place.php
class Place extends Eloquent {
public function photos() { return $this->morphMany('Image', 'of'); }
}
Here, a Person can upload one mugshotand many photos, while a Place can have many photos. The problem is that when I create a mugshoton a Person, it saves this into the imagestable in the database:
在这里，一个 Person 可以上传一个mugshot和多个photos，而一个 Place 可以上传多个photos。问题是，当我mugshot在 Person 上创建一个时，它会将其保存到images数据库中的表中：
id: 1
of_id: 1
of_type: Person
It doesn't store the fact that it's a mugshotand not a photo, so when I go to retrieve it, $person->mugshotmay sometimes return one of $person->photosand vice versa.
它不存储它是 amugshot而不是 a的事实photo，所以当我去检索它时，$person->mugshot有时可能会返回其中之一$person->photos，反之亦然。
Is there either (a) a better way to do this than creating 2 links on the same Model, or (b) a way to actually make this way work?
是否有 (a) 比在同一模型上创建 2 个链接更好的方法，或者 (b) 一种实际使这种方式起作用的方法？
回答by Sergiu Paraschiv
No built-in way right now. Maybe in Laravel 4.1 that's supposed to bring a complete rewrite of polymorphic relations.
目前没有内置方式。也许在 Laravel 4.1 中，这应该会带来对多态关系的完全重写。
Add a typeproperty to Image, then define whereconditions on the relations:
添加一个type属性到Image，然后定义where关系的条件：
public function mugshot() {
return $this->morphOne('Image', 'of')->where('type', 'mugshot');
}
public function photos() {
return $this->morphMany('Image', 'of')->where('type', 'photo');
}
Don't forget to set typeon Images you create.
Or, like I did bellow, hide that logic inside the model.
不要忘记type在Image您创建的 s上设置。或者，就像我在下面所做的那样，将该逻辑隐藏在模型中。
Here's my code (I'm using PHP 5.4 with short array notation):
这是我的代码（我使用的是带有短数组符号的 PHP 5.4）：
Image:
图片：
namespace SP\Models;
class Image extends BaseModel {
const MUGSHOT = 'mugshot';
const PHOTO = 'photo';
protected $hidden = ['type'];
public function of()
{
return $this->morphTo();
}
}
Person:
人：
namespace SP\Models;
use SP\Models\Image;
class Person extends BaseModel {
public function mugshot() {
return $this->morphOne('SP\Models\Image', 'of')
->where('type', Image::MUGSHOT);
}
public function photos() {
return $this->morphMany('SP\Models\Image', 'of')
->where('type', Image::PHOTO);
}
public function saveMugshot($image)
{
$image->type = Image::MUGSHOT;
$image->save();
$this->mugshot()->save($image);
}
public function savePhotos($images)
{
if(!is_array($images))
{
$images = [$images];
}
foreach($images as $image)
{
$image->type = Image::PHOTO;
$image->save();
$this->photos()->save($image);
}
}
}
Somewhere in a controller/service:
控制器/服务中的某处：
$person->savePhotos([$image1, $image2]);
