Yii get Model as parameter

Question

I am trying to modify 1 controller to work with 2 models. This is what I have done with loadModel function in my controller

public function loadModel($id, $_model)
{
    if (isset($_model)){
        $model=$_model::model()->findByPk($id); // syntax error, unexpected "::"
        if($model===null)
            throw new CHttpException(404,'The requested page does not exist.');
        return $model;
    } else {
        $model=Foods::model()->findByPk($id);
        if($model===null)
            throw new CHttpException(404,'The requested page does not exist.');
        return $model;
    }
}

As you can see I want to create optional parameter for this function where second parameter will be the Model. Can you help me to achieve this?

http://code2real.blogspot.in/2015/06/yii-11-get-current-model-name.html — Pupil, Jun 09 '15 at 10:19

score 2 · Accepted Answer · answered Sep 17 '13 at 09:42

2

You can't use string as class name to call static method. Just instantiate model and call findByPk:

if (isset($_model)){
    $model = new $_model;
    $model= $model->findByPk($id);

answered Sep 17 '13 at 09:42

score 1 · Answer 2 · answered Sep 17 '13 at 09:59

Perhaps it would be better

/**
* @var integer $id
* @var string $_model name model class
*/
       public function loadModel($id, $_model = 'Foods'){
                $model = new $_model;
                $model= $model->findByPk($id);
                if($model===null)
                    throw new CHttpException(404,'The requested page does not exist.');
                return $model;
        }

Yii get Model as parameter

2 Answers2