Yii 1.1: Update content in AJAX with renderPartial


The easiest way to update content in AJAX is to use the renderPartial method.

For this exemple I have three files: a controller (HelloWorldController.php) and two views (index.php and _ajaxContent.php)


class HelloWorldController extends CController
    public function actionIndex()
        $data = array();
        $data["myValue"] = "Content loaded";
        $this->render('index', $data);
    public function actionUpdateAjax()
        $data = array();
        $data["myValue"] = "Content updated in AJAX";
        $this->renderPartial('_ajaxContent', $data, false, true);

The actionIndex set myValue to "Content loaded" and this variable is passed to the view "index.php" and to "_ajaxContent.php"

Note: if using accessRules() in your controller file, you will need to modify accessRules() by adding the appropriate part of the function name to a rule set - in this case 'updateajax' like this:

array('allow',  // allow all users to perform 'index' and 'view' actions


<div id="data">
   <?php $this->renderPartial('_ajaxContent', array('myValue'=>$myValue)); ?>
<?php echo CHtml::ajaxButton ("Update data",
                              array('update' => '#data'));

The ajaxButton call "actionUpdateAjax" and the returned data are inserted in the div "data"


<?php echo $myValue ?>

Display $myValue

Now, run index.php?r=helloWorld


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Total 17 comments

#19867 report it
Burzum at 2016/05/27 02:36am
Redirection after validation

@kuncolaksmono if you want to perform a redirection after the validation, you can simply return a piece of javascript instead of html in "_ajaxContent.php" like:

<script type="text/javascript">
window.location = 'your-url';

Is that ok?

#19866 report it
kuncolaksmono at 2016/05/25 10:42pm
How to redirect if success submit form

I have a form, how to redirect if success submit form with model validation and ajax request ? thx

#16634 report it
Sergio_HW at 2014/03/13 06:00am
Update Modal Window

Hi, I need to update a modal window and show it when I click in the edit button.


public function actionUpdateAjax($id)
        $contratos = ZfContratos::model()->findByPk($id);
        $this->renderPartial('//ZfContratos/_form_update', array('model'=>$contratos), false, true);


<?php $this->beginWidget(
    array('id' => 'actualizar_contrato')
); ?>
    <div class="modal-header">
        <a class="close" data-dismiss="modal">&times;</a>
        <h4>Actualizar contrato</h4>
    <div class="modal-body">
        <?php $this->renderPartial('//ZfContratos/_form_update', array('model'=>$contrato));?>
    <div class="modal-footer">
        <?php $this->widget(
                'label' => 'CANCELAR',
                'url' => '#',
                'htmlOptions' => array('data-dismiss' => 'modal'),
        ); ?>
<?php $this->endWidget(); ?>

AND VIEW _contratos

At this moment I have this:

<?php echo CHtml::link('EDITAR', array('//ZfInmuebles/UpdateAjax', 'id'=>$data->zf_contrato_id), array('class'=>'btn', 'id'=>'vermas')); ?>

But I need that to be an ajaxbutton or ajaxlink, that refresh the div "actualizar_contrato" and show it.

#16351 report it
VincentM at 2014/02/13 11:02am
thx again :)

_edit : [Problem Solved] _

thx for your response, but I made something else. I send the form to the controller and then to the view with GET option. I don't know if it is possible with POST :/

If you wan't to keep helping me I posted that on the forum at this link :http://www.yiiframework.com/forum/index.php/topic/51416-how-to-send-a-cactiveform-widget/

But thanks again for your article, it helped me a lot :)

#16331 report it
Burzum at 2014/02/11 11:14am


I see your problem, in your _ajax something is missing:

<?php $form=$this->beginWidget('CActiveForm', array(
)); ?>
.... your form ...
<?php $this->endWidget(); ?>

That's it :)

#16327 report it
VincentM at 2014/02/11 06:21am
help me plz :)

thx great article but I have a problem.. :s (certainly a stupid newbie problem :p)

so in my view, I have a form with several option, then I put a comboBox with 2 choices. depending of this choice, I show different options in my form.

So I put this in my radioButton,

'onChange'=>CHtml::ajax(array(  'type'=>'POST',

then this in my controller

public function actionUpdateAjaxFormGestionnaire()
        $this->renderPartial('_ajaxContentFormGestionnaire', array(
                                                   ), false, true);

and this in my _ajax view file :

<td class="enableCreate">
                    <div class="row">
                        <?php echo $form->labelEx($eventForm,'Nom'); ?>
                        <?php echo $form->textField($eventForm,'Nom',array('size'=>50,'maxlength'=>50)); ?>
                        <?php echo $form->error($eventForm,'Nom'); ?>
                    <div class="row">
                        <?php echo $form->labelEx($eventForm,'Prenom'); ?>
                        <?php echo $form->textField($eventForm,'Prenom',array('size'=>50,'maxlength'=>50)); ?>
                        <?php echo $form->error($eventForm,'Prenom'); ?>

But the problem is that the $form widget is not initialized or passed or ... in the _ajax file. So I have this error :

Fatal error: Call to a member function labelEx() on a non-object in D:\WWW\VincentM\LayoutColloques\protected\views\admin\_ajaxContentFormGestionnaire.php on line 4

do you have any idea to help me? thx in advance.

#15825 report it
junaidatari at 2013/12/20 05:52am

Good work man, I like it!

#15823 report it
rajesh chaurasia at 2013/12/20 02:05am
nice article

its very helpfull article.

#13858 report it
Adil Shahzad at 2013/07/02 09:41am
Code for the Dropdown

If you need to updated the contents of a div through the onchange event of the dropdown then use the following code.

echo $form->dropDownListRow($model,'vehId', $listVeh , 
        'empty'=> 'Select Vehicle',
        'ajax' => array(
                        'type' => 'POST', 
                        'url' => CController::createUrl('UpdateAjax'),
                        'data'=> array('vehId'=>'js: $(this).val()'),  
#8655 report it
rix.rix. at 2012/06/17 10:41am
Instead of the ajaxButton you can also use an ajaxLink
echo CHtml::ajaxLink('clickMe', array('ajax'), array('update'=>'#ajax-results'));

This will call actionAjax() in the controller and the div with id 'ajax-results' will be updated.

Great tutorial by the way - thanks!

#7309 report it
Burzum at 2012/03/13 09:42am
You're welcome

@socialdev @vasireddy @nettrinity

You're welcome! I'm still woking with yii (of course, it's the best framework ever :) ), when I'll have more time I'll do more simple tutorials like this one :)

#7308 report it
socialdev at 2012/03/13 09:27am
The best shortest clear example!

Thank you my friend, thank you for sharing such a simple example which shows the whole concept of AJAX using Yii. I think this is the most beautiful and clear example I've ever seen together with some Android Java examples. Thank you!

#6614 report it
Fan_Of_Yii at 2012/01/21 05:29pm

Thanks for this wonderful tutorials .

Below is the code for radio button list

<?php echo CHtml::radioButtonList( 'cars','4', array(4 => 'GM', 5 => 'FORD'), array('separator' => '',
        'onChange'=>CHtml::ajax(array('type'=>'GET', 'url'=>array("editProfile/UpdateAjax"), 


#5583 report it
nettrinity at 2011/10/21 04:27pm

this is the kind of tutorials that I need!

#1942 report it
zaccaria at 2010/10/18 06:51am
process output

Pay attention that this fourth parameter set to true can create conflict with existing widgets outside the rendered subform.

This can happen because Yii uses a counter for generate the widget ID, and if you are not rendering all the widget (if there are some outside your panel) the new widget will have the same Id of the external.

A partial solution for this problem is to set a unique Id for each compoment inside the rendered panel, this solves many problem

#344 report it
gsatir at 2010/07/04 04:05pm
Notice that example has processOutput=true

The fourth parameter to renderPartial() is processOutput and it defaults to false. Note that it is set to true in actionUpdateAjax(). I've never paid any attention to this parameter until I tried to load a form through AJAX and the form has widgets on it. The widgets did not work because they weren't getting initialized by the usual call to .ready(). The solution seems to be processOutput=true. With that, the initialization code gets put into the HTML that is returned by the ajax call. My ajax-form-with-widgets now works.

#1058 report it
nexus246 at 2009/12/22 04:03am
No need for 2 actions

Instead of using 2 actions you can put all logic in one.

if(Yii::app()->request->isAjaxRequest) { $this->renderPartial('_ajaxContent',$data); } else { $this->render('index',$data); }

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  • Written by: Burzum
  • Updated by: hairylunch
  • Category: Tutorials
  • Yii Version: 1.1
  • Votes: +60 / -1
  • Viewed: 271,299 times
  • Created on: Sep 30, 2009
  • Last updated: Jul 10, 2012
  • Tags: AJAX