在laravel中通过url传递变量

Question

我对laravel相当陌生，而且我正努力使我的url格式正确。

它格式，

http://mysite/blog?category1 instead of http://mysite/blog/category1 

这是我使用的文件，有没有把路线进入BlogController

Route.php

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); 
}); 

Blogcontroller

方式

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)); 
    } 
} 

blog.index刀片

@foreach ($posts as $post) 

    <h2>{{ $post->id }}</h2> 
    <p>{{ $post->name }}</p> 
    <p>{{ $post->category }}</p> 
    <h2>{{ HTML::link(
    action('BlogController@index',array($post->category)), 
    $post->category)}} 


@endforeach 

Answer 1

而不是使用功能的回调为您Route::get使用一个控制器和一个动作：

Route::get('blog/{category}', 'BlogController@getCategory'); 

现在您的BlogController您可以创建功能。

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)); 
    } 

    /** 
    * Your new function. 
    */ 
    public function getCategory($category = null) 
    { 
     // get all the blog stuff from database 
     // if a category was passed, use that 
     // if no category, get all posts 
     if ($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); 
    } 
} 

更新：

要在视图中显示你的链接，你应该使用HTML::linkAction而不是HTML::link：

@foreach ($posts as $post) 

    <h2>{{ $post->id }}</h2> 
    <p>{{ $post->name }}</p> 
    <p>{{ $post->category }}</p> 
    {{ HTML::linkAction('BlogController@index', "Linkname", array('category' => $post->category)) }} 

@endforeach 

Answer 2

有如图所示，你试图使用可替代的.htaccess文档？ 在这里你去：

Options +FollowSymLinks 
RewriteEngine On 

RewriteCond %{REQUEST_FILENAME} !-d 
RewriteCond %{REQUEST_FILENAME} !-f 
RewriteRule^index.php [L] 

你需要把它放在你的应用程序的文件夹public。

这里是万一原来的.htaccess你没有它无论出于何种原因

<IfModule mod_rewrite.c> 
<IfModule mod_negotiation.c> 
    Options -MultiViews 
</IfModule> 

RewriteEngine On 

# Redirect Trailing Slashes... 
RewriteRule ^(.*)/$ /$1 [L,R=301] 

# Handle Front Controller... 
RewriteCond %{REQUEST_FILENAME} !-d 
RewriteCond %{REQUEST_FILENAME} !-f 
RewriteRule^index.php [L] 
</IfModule> 

Answer 3

我加入了一个新的途径：

Route::get('blog/{category}', ['as' => 'post.path', 'uses' => 'BlogController@getCategory']); 

，并添加新的链接进入index.blade：

<a href="{{ URL::route('post.path', [$post->category]) }}">{{ $post->category }}</a> 

Answer 4

routes.php

Route::get('category', 'CategoryController@indexExternal'); 

* .blade.php打印完成的URL

<a href="{{url('category/'.$category->id.'/subcategory')}}" class="btn btn-primary" >Ver más</a>