How to Retrieve Image Files from Database Using Yii Framework?

Question

Our Yii Framework application has the following defined as part of the UserProfileImages model:

public function getProfileImages($param, $user_id) {
        if(isset($param['select']) && $param['select']=='all'){
            $profile_images = UserProfileImages::model()->findAllByAttributes( array( 'user_id'=>$user_id) );

        } else {
            $profile_images = UserProfileImages::model()->findByAttributes( array( 'user_id'=>$user_id) );
        }
        return $profile_images;
    }

How would I wire up the above snippet to a widget in my view to return all the images for a given user?
Bonus Question: Which image rotator do you suggest to render the above?

BerndBrot · Accepted Answer · 2012-08-24 09:03:49Z

In your view file, add something like this, assuming that your controller specified $user_id:

$this->widget('UserProfileImagesWidget', array(
    "userProfileImages" => UserProfileImages::getProfileImages(
        array("select" => "all"), 
        $user_id
     ),
     "user_id" => $user_id
));

Depending on your MVC philosophy, you could also retrieve the userProfileImages data in the controller and pass that data to your view.

Define a widget like this:

class UserProfileImagesWidget extends CWidget {
    public $user_id;
    public $userProfileImages = array();

    public function run() {
        $this->render("userProfileImage");
    }
}

Finally, in the userProfileImages.php view file, you can do something like this:

if(!empty($this->userProfileImages)) {
    // Your display magic
    // You can access $this->user_id
}

As a side note: You might want to change the order of your parameters in getProfileImages. If $user_id is the first parameter, you can leave out $params completely in case you don't want to specify any.

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