We have a actionSearchType in our User Controller as follows:

```
public function actionSearchType()
{
        if (Yii::app()->user->isGuest == true)
            $this->render('login');
            else 
        $this->render('search_type');
    }

```

Our actionLogin in our User Controller is as follows:

```
public function actionLogin()
{
    $model= new Users();

    // if it is ajax validation request
    if(isset($_POST['ajax']))
    {
        echo CActiveForm::validate($model);
        Yii::app()->end();
    }
 $this->redirect(Yii::app()->user->returnUrl);
                    }
    }
    // display the login form
    $this->render('login',array('model'=>$model));
}

```

The goal is to ensure that only authenticated users can execute the options on the search type view. When I run this page, I receive an error stating `Undefined variable: model`.

A snippet of the login view is as follows:

```
<div class="form">
<?php $form=$this->beginWidget('CActiveForm', array(
'id'=>'login-form',
'enableClientValidation'=>true,
'clientOptions'=>array(
    'validateOnSubmit'=>true,
),
)); ?>

<p class="note">Fields with <span class="required">*</span> are required.</p>

<div class="row">
    <?php echo $form->labelEx($model,'username'); ?>
    <?php echo $form->textField($model,'username'); ?>
    <?php echo $form->error($model,'username'); ?>
</div>

```

What steps must be taken to remedy the above error and properly check to ensure we have an authenticated user?

**update**
I changed actionSearchType to render the Login Widget per below:

```
 public function actionSearchType()
{
        if (Yii::app()->user->isGuest)
            $this->widget('ext.LoginWidget');
            else 
        $this->render('search_type');
    }

```

This indeed resolved the error initially seen. A new problem is that there's no styling of the login widget when it renders. Should I echo my tags with appropriate stylesheet classes, or is there a bit more elegant way of doing that?