Laravel自定义路由模型绑定 - Laravel Custom Route Model Binding

Laravel Custom Route Model Binding

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'];
}

url是这样的:

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;但是我的问题是关于生成URL

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

简单地生成,localhost/页id

我知道我能做到:

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

但是如果可能的话，我更愿意做得干净一点。有人有什么建议吗?

自版本5以来，您现在可以在Model(通过UrlRoutable)上拥有getRouteKey()，您可以使用它来返回自定义路由键。比如:

class Page extends Node implements SluggableInterface
{
    //......
    public function getRouteKey() {
        return $this->generateURLString();
    }
}

使{{ route('page', $page) }}表现为您想要的。

文档

正如您所说，您可以执行{{URL::route('page',['page'=>$page->generateURLString()])}}，因为route('page',$page)将返回模式名称。

然后，我的建议是，如果你需要一些简洁，创建一个自定义函数来扩展Route类，或者只是将它声明为一个常规函数:

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

然后输入:

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