使用自定义路由模型绑定键输出 URL

问题描述 投票:0回答:2

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
    }
});

页面.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 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, 本地主机/页面 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。有人有推荐吗

php laravel-5 laravel-routing
2个回答
2
投票

从版本 5 开始,您现在可以在

getRouteKey()
(通过
Model
)上有一个
UrlRoutable
,您可以使用它来返回自定义路由密钥。 比如:

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

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

{{ route('page', $page) }}
表现得如你所愿。

文档


1
投票

正如你所说,你可以这样做

{{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,当您需要一些清洁剂时,创建一个扩展 Route 类的自定义函数,或者只是将其声明为常规函数:

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

然后就做:

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