Yii RenderPartial not working

Question

I am using renderParial() function to render views->subscriber->_form from views->layouts->main.php Problem i am getting is I am failed to insert data using this. Gii generated CRUD is working perfectly but when i render _form and want to insert data in database it's not working. Here is my code. main.php

<div class="col-lg-6" style="text-align:right;">
                     <span>Subscribe Newsletter</span>  
                      <div style="float:right;margin-left:10px;">
                            <?php
                                                   $model=new Subscription();

                        $this->renderPartial("/subscription/_form",array('model'=>$model));?>

                  </div>

and _form.php

<?php
/* @var $this SubscriptionController */
/* @var $model Subscription */
/* @var $form CActiveForm */
?>

<div class="form">

<?php $form=$this->beginWidget('CActiveForm', array(
    'id'=>'subscription-form',
    // Please note: When you enable ajax validation, make sure the corresponding
    // controller action is handling ajax validation correctly.
    // There is a call to performAjaxValidation() commented in generated controller code.
    // See class documentation of CActiveForm for details on this.
    'enableAjaxValidation'=>false,
)); ?>



    <?php echo $form->errorSummary($model); ?>


    <div class="row">

        <?php echo $form->textField($model,'email',array('size'=>30,'maxlength'=>30,'placeholder'=>'Email Address')); ?>
        <?php echo $form->error($model,'email'); ?>
        <?php echo CHtml::submitButton($model->isNewRecord ? 'Subscribe' : 'Save',array('class'=>'btn btn-xs btn-success subscribeBtn')); ?>
</div>

<?php $this->endWidget(); ?>

</div>

This goes against the whole MVC pattern. You can better create a widget with this form and include it in your main layout and you model. — Michiel, Feb 06 '14 at 10:35
agree with @Michiel you shouldn't put this type of thing in your layout. — adamS, Feb 06 '14 at 10:37
Where is your $model coming from, what controller/action are you declaring it in? Creating a widget would allow you to specify which model (in the widget class) you are trying to save the form data with. http://www.yiiframework.com/doc/guide/1.1/en/basics.view#widget — adamS, Feb 06 '14 at 10:44
@adamS I am declaring $model in my main.php file and sending $model as a parameter in renderPartial(). — Mushahid Hussain, Feb 06 '14 at 10:48
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/46919/discussion-between-adams-and-james) — adamS, Feb 06 '14 at 11:05

deacs · Accepted Answer · 2014-02-06T11:44:24.703

As adamS and Michiel have commented, if you want to put a form or data in your main.php layout file, you should use a widget.

To create a widget, you need to do the following:

1: Create a php file in your /protected/components/ dir, something like SubscriptionWidget.php

2: Create a dir views in your components dir

3: Create your view .php file in your /protected/components/views/, something like subscriptionWidget.php

4: Put the following code in your SubscriptionWidget.php file:

<?php 
class SubscriptionWidget extends CWidget
{
    public function init()
    {

    }

    public function run()
    {
        $model = new SubscriptionForm;
        if(isset($_POST['SubscriptionForm']))
        {
            // proces the data
        }
        $this->render('subscriptionWidget', array('model'=>$model));
    }
}
?>

Your widget is done. All you need to do now is call it in your main.php layout file, like so:

<!doctype html>
<html lang="en">
...
<?php $this->widget('SubscriptionWidget'); ?>
...
</html>

Also, don't forget to put the form in your newly created view file.

Hope this helps.

score 0 · Answer 2 · answered Feb 06 '14 at 10:50

0

Try adding one more slash

$this->renderPartial("//subscription/_form",array('model'=>$model));

answered Feb 06 '14 at 10:50

Let me see

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Yii RenderPartial not working

2 Answers2