There are two possible ways: instantiating the M [...]]]> I don’t know if the following is of any practical use, I was just curious whether it is possible to use a table without creating a corresponding model class for it. And yes, it is possible.

There are two possible ways: instantiating the Model class manually, or using the ClassRegistry. First the manual way, which requires a bit more work:

class ExamplesController extends AppController {
    public $uses = array();
	
    public function index() {
        App::import('Core', 'Model');
        $this->Month = new Model(false, 'months');
        $this->Month->alias = 'Month';
        $this->set('months', $this->Month->find('all'));
    }
}

In the example above we create an instance of the Model class which uses the “months” table (and we don’t set an ID, hence the “false” in the constructor). Then we have to specify the alias property (and not the name property as you may guess). Without defining this property, the result of the find statement would be as shown below (notice the “Model” key):

Array
(
    [0] => Array
        (
            [Model] => Array
                (
                    [id] => 1
                    [name] => January
                )
        )
    ...
)

The second approach uses the ClassRegistry to instantiate the Model class. The example from above can now be shortened to:

class ExamplesController extends AppController {
    public $uses = array();
	
    public function index() {
        $this->Month = ClassRegistry::init(array('class' => 'Model', 'table' => 'months', 'alias' => 'Month'));
        $this->set('months', $this->Month->find('all'));
    }
}

As I said at the beginning of this article, this was just an experiment, but maybe you come up with a practical use case for it ;-)