CakePHP and conditions - php

Question

So in my controller MenuController.php I have the following code:
class MenuController extends AppController {
public $helpers = array('Html', 'Form');
public function index() {
$this->set('menus', $this->Menu->find('all'));
$userSpecific = $this->Menu->find('all', array(
'conditions' => array('Menu.user_id' => '20')
));
}
}
and in my view, I am doing the following:
<?php foreach ($menus as $menu): ?>
<?php echo $menu['Menu']['id']; ?>
<?php echo $menu['Menu']['user_id']; ?>
<?php endforeach; ?>
update
To better understand this in my browser I changed my view to the following:
<?php foreach ($menus as $menu): ?>
<p>Menu id <?php echo $menu['Menu']['id']; ?> is for user: <?php echo $menu['Menu']['user_id']; ?></p>
<?php endforeach; ?>
end update
Now in the view, it is currently using the $this and returning all values from the database table, How do I change the view to use $userSpecific rather than $this I managed to get this far (making the $userSpecific part) by using the cookbook but I could only find the controller side and not the view side. I'm sorry if it's a bad question, just trying to learn.

Answer 1

You need to send the data to the view from the controller at the end of the index() function.
You can do it like this:
$this->set('userSpecific', $userSpecific);
or like this (my preferred way)
$this->set(compact('userSpecific');
Once you've done this, you can then modify your view to show the user specific fields as shown:
<?php foreach ($userSpecific as $menu): ?>
<?php echo $menu['Menu']['id']; ?>
<?php echo $menu['Menu']['user_id']; ?>
<?php endforeach; ?>