Displaying Content with laravel template without blade

Question

I have this controller method

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

<div id = "content">
    <?php echo Section::yield('content'); ?>
</div>

And this is my test.php

echo $a;
echo '<br>';
echo $b;
echo 'this is content';

When i access this

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.

score 4 · Accepted Answer · answered Mar 08 '13 at 04:49

When you nest a view within another it's content is defined as a simple variable. So, simply output it:

<?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:

// 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>

Great! it is too simple i never paid attention this could be too simple. Thanks! — Muhammad Raheel, Mar 08 '13 at 06:44

score 2 · Answer 2 · edited May 23 '17 at 12:16

I think you need render for it, not sure, maybe partial loading:

<div class="content">
    <?php echo render('content.test'); ?>
</div>

Look this sample for nesting views: 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

Displaying Content with laravel template without blade

2 Answers2