使用没有刀片的 Laravel 模板显示内容
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Displaying Content with laravel template without blade
提问by Muhammad Raheel
I have this controller method
我有这个控制器方法
class Account_Controller extends Base_Controller
{
public $layout = 'layouts.default';
public function action_index($a,$b)
{
$data['a'] = $a;
$data['b'] = $b;
$this->layout->nest('content', 'test',$data);
}
}
And this is my layout
这是我的布局
<div id = "content">
<?php echo Section::yield('content'); ?>
</div>
And this is my test.php
这是我的 test.php
echo $a;
echo '<br>';
echo $b;
echo 'this is content';
When i access this
当我访问这个
http://localhost/myproject/public/account/index/name/email
I get my layout loaded but test.php is not loaded. How can i load content in my template. I dont want to use blade.
我加载了我的布局，但未加载 test.php。如何在我的模板中加载内容。我不想用刀片。
回答by vFragosop
When you nest a view within another it's content is defined as a simple variable. So, simply output it:
当您将一个视图嵌套在另一个视图中时，它的内容被定义为一个简单的变量。所以，简单地输出它：
<?php echo $content ?>
Section is used when you need to change something on your layout (or any parent view really) from within the child view. For instance:
当您需要从子视图中更改布局（或任何父视图）上的某些内容时，将使用部分。例如：
// on layout.php
<title><?php echo Section::yield('title') ?></title>
// on test.php
<?php Section::start('title'); ?>
My Incredible Test Page
<?php Section::stop(); ?>
<div class="test_page">
...
</div>
回答by Marin Sagovac
I think you need render for it, not sure, maybe partial loading:
我认为你需要渲染它，不确定，也许是部分加载：
<div class="content">
<?php echo render('content.test'); ?>
</div>
Look this sample for nesting views: http://laravel.com/docs/views#nesting-views
查看此示例以获取嵌套视图：http: //laravel.com/docs/views#nesting-views
public function action_dostuff()
{
$view = View::make('controller.account');
// try dump var to grab view var_dump($view);
var_dump($view);
$view->test = 'some value';
return $view;
}
Or use instead blade: Templating in Laravel
或者使用刀片：Laravel 中的模板
