Yii Framework: Undefined variable model when checking if guest - php

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,
));
}

Answer 2

Instead of rendering the view redirect to the user login page / action so you don't have to recreate it.
$this->redirect('login');

Answer 3

Somewhere in search_type you are referencing the variable $model which you do not hand over to the render() function. You need to define that variable otherwise the view will create an Exception.
I don't know which Model/Class your search_type view is expecting but you will need to initialize it before you hand it over to the view like this:
$this->render('search_type',array(
'model' => $model,
));
Here a good read about this topic: Understanding the view rendering flow