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Laravel 4,查看带参数的composer?

转载 作者:行者123 更新时间:2023-12-03 15:05:14 25 4
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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')

我的问题是,我怎么能转移:
$data["posts_related"] = Blog::posts_related($post->category_id,5);

对于 View Composer ,因为我似乎无法将参数传递给 View Composer ,但我不能确定。

我感谢任何帮助!

最佳答案

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);
});

现在到处都可以使用:
 View::make('sidebars.related')->with(array('category_id' => 6));

关于Laravel 4,查看带参数的composer?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16178549/

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