Difference between #5 and #4 of Instantiate a child class that depends a type field from parent model class (CActiveRecord)

Instantiate a child class that depends a type field from parent model class (CActiveRecord)
activerecord, CActiveRecord, factory, instance of child model, active
**[EN]** [[ES]](http://www.yiiframework.com/wiki/414/)

Sometimes we have an scenario like this:

class AnimalType extends CActiveRecord {} // Table with the "types" of
Animals available

class Animal extends CActiveRecord { // Parent model class
  // Model based on a table with a field: type_id
  // that stores the type of the Animal

class Cat extends Animal {} // Child class

class Dog extends Animal {}

And we'd like to do something like:

// This statement will return a "Cat" instance
$someAnimal = Animal::model()->findByPk(1);

Since Yii implements a factory design model, it's very easy to implement a
simple and quick solution for this.

**Step 1** (and unique)
Override **instantiate()** function.
_I included the PhpDoc comments present in CActiveRecord_

     * Creates an active record instance.
     * This method is called by {@link populateRecord} and {@link
     * You may override this method if the instance being created
     * depends the attributes that are to be populated to the record.
     * For example, by creating a record based on the value of a column,
     * you may implement the so-called single-table inheritance mapping.
     * @param array $attributes list of attribute values for the active records.
     * @return MyClass the active record
    protected function instantiate($attributes)
        if (!isset($attributes['type_id'])) { // When creating and $type_id has
not defined yet
            return parent::instantiate($attributes);
        $classRecord = AnimalType::model()->findByPk($attributes['type_id']);
// Load the "Type" class
        if($classif($classRecord === null)
            throw new CException('Type not found');
        $className = $classRecord->name; // Let's assume that name is the
field storing the name of the class
        $model = new $className(null); // Null param needed for populating the
        return $model;

Very easy, right.
Enjoy and share :)
_If you find some grammar, spelling or language mistake, feel free to fix it or
report it to me. English is not my native language_