如何在Laravel 4路由URL中同时使用ID和slug？(资源/ ID /段塞)

Question

我想在我的文章路线中同时使用ID和slug.而不是/articles/ID我想要的/articles/ID/slug.

我实际上并不需要slug变量; 它只是为了使URL更具可读性和SEO,所以我将ID用作检索文章的标识符.

如果/articles/ID输入了URL ,我想重定向到/articles/ID/slug.必须有一个例外,/articles/ID/edit因为这会打开用于编辑文章的表单.

我用谷歌搜索并查看了这个网站,但我只找到了用slug替换ID的例子,不包括两者.

我怎样才能做到这一点？我可以使用URL该类获取文章的完整URL(/articles/ID/slug)吗？

当前路由配置:

Route::resource('articles', 'ArticlesController');

Answer 1

所以,这是我最终做的事情:

routes.php,为show和创建了自定义路由edit.使用剩下的资源:

Route::pattern('id', '[0-9]+');

Route::get('articles/{id}/{slug?}', ['as' => 'articles.show', 'uses' =>   'ArticlesController@show']);
Route::get('articles/edit/{id}', ['as' => 'articles.edit', 'uses' => 'ArticlesController@edit']);
Route::resource('articles', 'ArticlesController', ['except' => ['show', 'edit']]);

Controller,添加了一个slug带默认值的输入参数.如果slug缺失或不正确,则重定向请求,因此如果标题被更改,它将重定向并返回HTTP 301(永久移动):

public function show($id, $slug = null)
{
    $post = Article::findOrFail($id);

    if ($slug != Str::slug($post->title))
        return Redirect::route('articles.show', array('id' => $post->id, 'slug' => Str::slug($post->title)), 301);

    return View::make('articles.show', [
        'article' => Article::with('writer')->findOrFail($id)
    ]);
}

查看演示者,我最初在我的模型类中有一些东西.但在此答案的基础上将其移至视图演示者类:https://stackoverflow.com/a/25577174/3903565,安装并使用此:https://github.com/laracasts/Presenter

public function url()
{
    return URL::route('articles.show', array('id' => $this->id, 'slug' => Str::slug($this->title)));
}

public function stump()
{
    return Str::limit($this->content, 500);
}

查看,从视图演示者处获取网址:

@foreach($articles as $article)
    <article>
        <h3>{{ HTML::link($article->present()->url, $article->title) }} <small>by {{ $article->writer->name }}</small></h3>
        <div class="body">
            <p>{{ $article->present()->stump }}</p>
            <p><a href="{{ $article->present()->url }}"><button type="button" class="btn btn-default btn-sm">Read more...</button></a></p>
        </div>
    </article>
@endforeach