Yii php ajax Undefined variable: model

Question

I'm still struggling with ajax form and Yii. I got the error Undefined variable: model. My form is renderPartial by another view renderPartial itself by another view.

Here the details

in my POST Controller

public function actionViewComment()
    {   
        $post=$this->loadModel();
        $comment=new Comment;
        $this->renderPartial('_viewComment',array(
            'model'=>$post,
            'comment'=>$comment,
        ));

}

_viewcomment.php call the commment/_form.php

<?php $this->renderPartial('/comment/_form',array(
            'model'=>$comment,
        )); ?>

in the _form.php , the variable $model is accessible there is a button that call a jquery function send() when it is clicked

function send() {

   var data=$("#comment-form").serialize();


  $.ajax({
   type: 'POST',
    url: '<?php echo Yii::app()->createAbsoluteUrl("post/Ajax"); ?>',
   data:data,
success:function(data){
                alert("Data Saved"); 
              },
   error: function(data) { // if error occured
         alert("Error occured.please try again");
         alert(data);
    },

  dataType:'html'
  });

}

The actionAjax of the post controller contains

public function actionAjax()
    {
if(isset($_POST['Comment']))
        {
           if($model->validate())
            {
               $model->save();
               return;
            }
        } 
     }

In action Ajax , I got the error Undefined variable: model. I dont understand why?

If I use $model=new Comment; the data is not saved because $model is empty

Can you explain?

by adding

$model=new Comment;
$model->attributes = $_POST["Comment"];

I go the following error

DbCommand failed to execute the SQL statement: SQLSTATE[23000]: Integrity constraint violation: 19 tbl_comment.status may not be NULL. The SQL statement executed was: INSERT INTO 'tbl_comment' ("author", "email", "url", "content", "create_time") VALUES (:yp0, :yp1, :yp2, :yp3, :yp4) (C:\wamp\www\yii\framework\db\CDbCommand.php:358)</p><pre>#0 C:\wamp\www\yii\framework\db\ar\CActiveRecord.php(1077): CDbCommand->execute()

Thank you in advance for your help.

Kumar V · Accepted Answer · 2014-02-07 17:27:48Z

1

In your actionAjax, you have to create model object for Comment. You didn't create $mode but you are calling their functions.

public function actionAjax()
    {
        if(isset($_POST['Comment']))
        {
           $model=new Comment;
           $model->attributes = $_POST["Comment"];
           if($model->validate())
            {
               $model->save();
               echo "saved";
            }
            else
            {
               echo "Failed";
            }
        } 
     }

edited Feb 7, 2014 at 17:27

answered Feb 7, 2014 at 17:08

Kumar V

8,8289 gold badges44 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

12 Comments

klark Over a year ago

I did already but in this case, the data is not saved. I looked to the firebug and data sent is empty. @kumar_v

Kumar V Over a year ago

So your post is empty?

klark Over a year ago

No , it is not. I can see in firebug that $_POST is equal to my data entered @kumar_v

klark Over a year ago

The response is stil empty in firebug. @kumar_v

Kumar V Over a year ago

try updated code. however you will get some response from ajax. so if you get failed, then validation failed. then check your model validation and data

|

Collectives™ on Stack Overflow

Yii php ajax Undefined variable: model

1 Answer 1

12 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

12 Comments

Your Answer

Sign up or log in

Post as a guest

Related