Question

I am wondering whether it is possible to create a hasMany Relationship in a Model which makes use of the ID of the logged in user.

Example:
One tip has many votings (from different users). This is easy:

public $hasMany = array(
   'TipVoting' => array(
        'className' => 'TipVoting',
        'foreignKey' => 'tip_id',
        'dependent' => false,
        'conditions' => '',
        'fields' => '',
        'order' => '',
        'limit' => '',
        'offset' => '',
        'exclusive' => '',
        'finderQuery' => '',
        'counterQuery' => ''
    ));

Now I want to get all Tips but want to know whether the logged in user hast voted this Tip already. So I would like to create a Realtionship which looks like this:

public $hasMany = array(
   'MyTipVoting' => array(
        'className' => 'TipVoting',
        'foreignKey' => 'tip_id',
        'dependent' => false,
        'conditions' => array('TipVoting.user_id' => $loggedinUser_id),
        'fields' => '',
        'order' => '',
        'limit' => '',
        'offset' => '',
        'exclusive' => '',
        'finderQuery' => '',
        'counterQuery' => ''
    )
);

How do I get the $loggedinUser_id into this Model? Or is this a bad way of implementing this issue? Should I rather go for a seperate find on the TipVoting Model?

Any ideas?

Thanks!

Answer 1

Use the containable behavior. http://book.cakephp.org/view/474/Containable

You don’t want to define a different relationship, you just want to be able to filter by associated criteria. Containable will allow you to do this.

With just the TipVoting relationship, in your model do this:

var $actsAs = array( 'Containable' );

in your controller action do this:

$this->TipVoting->find( 'all', array( 
 'contain' => array( 
   'Vote' => array( 
    'conditions' => array( 
      'TipVoting.user_id' => $this->Auth->User('id') ), 
    'order' => 'TipVoting.id DESC' ) ) );