Question

我对laravel相当新，而且我正努力让我的网址格式正确。

格式为

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

这些是我正在使用的文件，是否有办法将路径放入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

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 blade

@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

routes.php文件

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>

Answer 2

不使用函数作为Route::get的回调，而是使用控制器和操作：

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

现在，在您的BlogController中，您可以创建自己的功能。

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);
    }
}

<强>更新

要在视图中显示您的链接，您应使用HTML::linkAction代替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 3

您是否尝试过使用文档中显示的替代.htaccess？你走了：

Options +FollowSymLinks
RewriteEngine On

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

您需要将其放在应用程序的public文件夹中。

这是最初的.htaccess，以防你出于某种原因没有它

<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 4

我添加了一条新路线：

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

并在 index.blade：

中添加了一个新链接

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

在laravel中通过url传递变量

4 个答案: