Php laravel中的动态url

我是一个初学者，所以请原谅我，如果这是一个简单的问题

我试图通过我的url传递一个db变量，例如blog？=category 它显示所有结果，但当我将url更改为blog/category1时，它不会显示任何内容

数据库字段：id、类别、名称 1类别1试验

这是我的路线：

Route::get( '/blog', 'BlogController@index' );
Route::get('/', 'HomeController@index');
 Route::get('blog/{category}', function($category = null)
{
// get all the blog stuff from database
// if a category was passed, use that
// if no category, get all posts
if ($category)
$posts = Post::where('category', '=', $category);
else
$posts = Post::all();

// show the view with blog posts (app/views/blog.blade.php)
return View::make('blog.index')
->with('posts', $posts);
});

博客控制器

class BlogController extends BaseController {


    public function index()
    {
        // get the posts from the database by asking the Active Record for "all"
        $posts = Post::all();

        // and create a view which we return - note dot syntax to go into folder
        return View::make('blog.index', array('posts' => $posts));
    }
}

这是我的视图布局：

@extends('base')

@section('content')
@foreach ($posts as $post)
<h2>{{ $post->id }}</h2>
<p>{{ $post->name }}</p>
@endforeach
@stop

@extends（'base'））
@节（“内容”）
@foreach（$posts作为$post）
{{$post->id}
{{$post->name}

@endforeach
@停止

---

谢谢

您需要在雄辩的查询之后添加

->get（）

，以获取结果

Route::get('blog/{category}', function($category = null)
{
    // get all the blog stuff from database
    // if a category was passed, use that
    // if no category, get all posts
    if ($category)
        $posts = Post::where('category', '=', $category)->get();
    else
        $posts = Post::all();

    // show the view with blog posts (app/views/blog.blade.php)
    return View::make('blog.index')->with('posts', $posts);
});

---

很高兴我能帮忙。把这个答案标记为解决方案。