Question

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?

Answer 1

public function actionSearchType() {
  if (Yii::app()->user->isGuest)
    $this->redirect('/user/login');

  $this->render('search_type');
}

Notes:

• to do something when user is guest, simply use if(Yii::app()->user->isGuest) { statement }

• to do something when user is logged in, simply use if(!Yii::app()->user->isGuest) { statement }

in the second code, public function actionLogin(), I think you have 2 more closing curly brackets than needed. Anyway, the login action should look like this:

public function actionLogin() {
    $formModel = new Login_Form; // Login_Form.php should be in models folder
    if (isset($_POST['Login_Form'])) {
        $formModel->attributes = $_POST['Login_Form'];
        if ($formModel->validate() && $formModel->login()) {
            $this->redirect('/'); // replace / with stuff like Yii::app()->user->returnUrl
        }
    }
    $this->render('login', array(
        'formModel'=>$formModel,
    ));
}