Hello
With ajaxlink i’m picking which item i want to proceed with…
I then want the controller to update my div id with a linkbutton.
My code:
echo CHtml::ajaxLink(
CHtml::image($data->imageUrl, $data->step1Name, array("style"=>"width: 80px; height: 70px; border: 1px;")),
Yii::app()->createUrl( '/fs/ajaxRequest' ),
array(
'type'=>'POST',
'update'=>'#proceed_button',
'data'=>array(
'step'=>'step1',
'id'=>$data->step1Id
)
),
array(
'href' => Yii::app()->createUrl( '/fs/ajaxRequest' ),
'class' => 'link',
'id'=>uniqid(),
'title'=>$data->step1Name,
)
);
<div id="proceed_button"><?php echo CHtml::submitButton('Proceed', array('class'=>'btn','disabled'=>'disabled')); ?></div>
My controller then looks like this:
public function actionAjaxRequest() {
if($_POST['step'] == 'step1'){
Yii::app()->session['step1'] = $_POST['id'];
echo CHtml::linkButton('Proceed', array('submit'=>array('goto'),'params'=>array('redirect'=>'step2','id'=>$_POST['id']),'class'=>'btn link8'));
}
Yii::app()->end();
}
This way, it will only echo the linkbutton, but not the actual jquery where the actual link is posted…
How can i accomplish this?
(i dont want to use a normal "link" if i can be without…
Thanks
// Casper