Laravel 自定义路由模型绑定

Question

我有以下设置：

routes.php

Route::get('{page?}', [
    'uses'=>'PageController@getPage',
    'as'=>'page'
])->where('page', '(.*)?');

RouteServiceProvider.php

$router->bind('page', function($value, $route)
{
    if($value == "/"){ $value = "home"; };
    $explodedPage = explode("/",$value);
    $page = Page::findBySlug(last($explodedPage));
    if(!isset($page)){
        \App::abort(404);
    }
    $ancestors = $page->ancestorsAndSelf()->get();
    $sections=array();
    foreach($ancestors as $ancestor)
    {
        $sections[]=$ancestor->slug;
    }
    if(implode("/",$sections)==$value){
        return $page;
    }else{
        return $page;
        //Else Redirect
    }
});

Page.php

use Baum\Node;
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;
use URL;
use Venturecraft\Revisionable\RevisionableTrait;
use Illuminate\Database\Eloquent\SoftDeletes;
use Cviebrock\EloquentSluggable\SluggableInterface;
use Cviebrock\EloquentSluggable\SluggableTrait;

class Page extends Node implements SluggableInterface
{

    use RevisionableTrait, SoftDeletes, SluggableTrait;

    protected $sluggable = array(
        'build_from' => 'title',
        'save_to'    => 'slug',
    );

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $fillable = ['title', 'description', 'content', 'owner_id', 'system', 'status'];

    /**
     * The attributes excluded from the model's JSON form.
     *
     * @var array
     */
    protected $hidden = ['parent_id','lft','rgt','depth'];

    /**
     * The attributes excluded from revision
     *
     * @var array
     */
    protected $dontKeepRevisionOf = ['updater_id','parent_id','lft','rgt','depth'];

}

URLS 看起来像这样：

localhost/ (uses pre-defined slug)  
localhost/page-slug  
localhost/parent-slug/page-slug  
localhost/parent-parent-slug/parent-slug/page-slug  
Etc...

检索页面工作正常；但我的问题是关于生成 URL

{{URL::route('page',$page)}}

简单生成， localhost/page-id

我知道我能做到：

{{URL::route('page',['page'=>$page->generateURLString()])}}

但如果可能的话，我更愿意做这个清洁工。有人有什么建议吗？

Answer 1

如您所说，您可以{{URL::route('page',['page'=>$page->generateURLString()])}}因为route('page',$page)将return模式名称。

然后，我的建议是，因为你需要一些更清洁的东西，所以创建一个自定义函数来扩展路由 class 或者只是将其声明为常规函数：

public function page($bind){
    return route('page', ['page' => $bind]);
}

然后就这样做：

{{ page($page->generateURLString()) }}

Answer 2

从版本 5 开始，您现在可以在 Model（通过 UrlRoutable）上拥有一个 getRouteKey()，您可以将其用于 return 您的自定义路由键。类似于：

class Page extends Node implements SluggableInterface
{
    //......

    public function getRouteKey() {
        return $this->generateURLString();
    }
}

让{{ route('page', $page) }}如你所愿。

Docs

Laravel 自定义路由模型绑定

Laravel Custom Route Model Binding

php

laravel-routing

laravel-5