Laravel 4,用参数查看作曲家?

use*_*178 4 laravel laravel-4

假设我有这个:

//controllers/BlogController.php
$data["post"] = $post = Blog::recent_post();
$data["posts_related"] = Blog::posts_related($post->category_id,5);
return View::make('blog.home', $data);

//views/sidebars/related.blade.php
@foreach($posts_related as $r)
<p>{{ $r->name }}</p>
@endforeach

//views/blog/home.blade.php
@include('sidebars.related')
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我的问题是,我怎么能转移:

$data["posts_related"] = Blog::posts_related($post->category_id,5);
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对于视图作曲家来说,因为我似乎无法将参数传递给视图作曲家,但我无法确定.

我感谢任何帮助!

Mil*_*Rey 29

在View Composer中,您可以检索传递给View的数据.

View::composer('sidebars.related', function($view)
{
   //$data contains category_id
   $viewdata= $view->getData();

   //Retrive based on category_id
   $posts_related = Blog::posts_related($viewdata['category_id'],5);

   //pass related post to view
   $view->with('posts_related', $posts_related);
});
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现在到处都可以使用:

 View::make('sidebars.related')->with(array('category_id' => 6));
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