Question

I’m using CakePHP 2.0 and I’m trying to understand how I can echo dynamic contents if the user is logged in the application.

In the view I would like to echo a menu to login or logout an user, how can I do that?

// I'm in the default template view
if (!AuthComponent::loggedIn()) {
    $menu  = $this->Html->link('Login', array('controller' => 'users', 'action' => 'login'));
    $menu .= $this->Html->link('Register',  array('controller' => 'users',  'action' => 'register'));
} else {
    $menu  = $this->Html->link('Home', array('controller' => 'users', 'action' => AuthComponent::user('id'), AuthComponent::user('username')));
    $menu .= $this->Html->link('Logout', array('controller' => 'users', 'action' => 'logout'));
}
echo $menu;

I thought something like this but I’ve read It breaks the MVC rules.

How should I do things like that in CakePHP?
Does exists some example online?

Answer 1

You could set if they’re logged in or not in a controller then use that element accordingly.

In your controller:

function beforeFilter() {
    if($this->Auth->loggedIn()) {
        $userBar = 'memberBar';
    } else {
        $userBar = 'guestBar';
    }
    $this->set('userBar', $userBar);
}

In your layout:

<?php echo $this->element($userBar); ?>

Then have a memberBar element and a guestBar element:

echo $this->Html->link('Home', array('controller' => 'users', 'action' => AuthComponent::user('id'), AuthComponent::user('username')));
echo $this->Html->link('Logout', array('controller' => 'users', 'action' => 'logout'));

You could pass the AuthComponent data to the element to avoid using the object in your layout.