Yii validate dropdown onchange

Question

I am trying to validate the dropdownlist as the value of dropdownlist changes. I want to check is there an in the the table already of the selected job status. Below is my code:

<script>
     function validate_dropdown(id)
    {
        alert("Selected id = "+id);
        //var msg = <?php echo NotificationRules::model()->validate_job($_POST['id']);?>
        //alert("Message from model func = "+msg);
    }
</script>

<?php
  echo $form->dropDownList($model, 'job_status_id', $jobstatuslist ,
            array('empty'=>'Please Select job status (required)', 'onchange'=>'js:validate_dropdown(this.value)')
        );


?>

I am trying to pass js variable id to php function and send back a message if there is already an entry for the selected job status. I am able to get selected value in js function validate_dropdown(), but not able to proceed further.Anybody pls help.......

Answer 1

I solved the problem by making an AJAX call...

Final working code:

Code of dropdown in view

<?php
   echo $form->dropDownList($model, 'job_status_id', $jobstatuslist ,
             array(//AJAX CALL.
                    'prompt' => 'Please Select job status (required)',
                    'value' => '0',
                    'ajax'  => array(
                            'type'  => 'POST',
                            'url' => CController::createUrl('NotificationRules/notificationPresent/'),
                            'data' => array("job_id" => "js:this.value"),
                            'success'=> 'function(data) {
                                    if(data == 1)
                                    {
                                        alert("Rule is already present for this status, Please update existing rule.");
                                    }
                                }',
                            'error'=> 'function(){alert("AJAX call error..!!!!!!!!!!");}',
                    )//end of ajax array().
        ));

?>

Code in controller(action)

<?php
   public function actionNotificationPresent()
   {
    if (Yii::app()->request->isAjaxRequest)
    {
        //pick off the parameter value
        $job_id = Yii::app()->request->getParam( 'job_id' );
        if($job_id != '')
        {
            //echo "Id is received is ".$job_id;

            $rulesModel = NotificationRules::model()->findAllByAttributes(array('job_status_id'=>$job_id));

            if(count($rulesModel))
                echo 1;
            else 
                echo 0;

        }//end of if(job_id)
        else
        {
            //echo "Id id not received";
            echo 0;

        }

    }//end of if(AjaxRequest).

}//end of NotificationPresent().

?>

Now i am getting an alert if there is any rule already with selected job status.

Answer 2

Check this bellow example. In this i'm displaying all the users in a drop down list. I'm keeping user id as option value and username as option label.

User Table:

    id  username    
    ------------------
    1   Heraman
    2   Dileep  
    3   Rakesh
    4   Kumar   


    <?php
    $list=CHtml::listData(User::model()->findAll(), 'id', 'username');       
    echo CHtml::dropDownList('username', $models->username, $list, array('empty' => '---Select User---','onchange'=>'alert(this.value)'));
    ?>

In you case, you can use

 'onchange'=>'validate_dropdown(this.value)

  //Your script
  <script>
        function validate_dropdown(id)
        {
            alert("Selected id = "+id);        
        }
    </script>