Question

I have just recently started working with Laravel. Great framework so far!
However I have a question.

I am using a layout template like this:
public $layout = 'layouts.private';

This is set in my Base_Controller:

public function __construct(){

    //Styles
    Asset::add('reset', 'css/reset.css');
    Asset::add('main', 'css/main.css');

    //Scripts
    Asset::add('jQuery', 'http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js');

    //Switch layout template according to the users auth credentials.
    if (Auth::check()) {
        $this -> layout = 'layouts.private';
    } else {
        $this -> layout = 'layouts.public';
    }

    parent::__construct();

}

However I get an error exception now when I try to access functions in my diffrent controllers, which should not call any view, i.e. when a user is going to login:

class Login_Controller extends Base_Controller {

public $restful = true;

public function post_index()
{

    $user = new User();
    $credentials = array('username' => Input::get('email'), 'password' => Input::get('password'));

    if (Auth::attempt($credentials))
    {

    } else {

    }

}

}

The error I get, is that I do not set the content of the different variables in my public $layout. But since no view is needed in this function, how do I tell Laravel not to include the layout in this function?

The best solution that I my self have come a cross (don’t know if this is a bad way?) is to unset($this -> layout); from function post_index()…

To sum up my question: how do I tell Laravel not to include public $layout in certain functions, where a view is not needed?

Thanks in advance,
fischer

Answer 1

You need to set $this->layout = null on any function you don’t want to render the view.