Julia

Thanks Edward for your suggestion. I have learnt something that I did not know even though it is not the solution to the problem. I sought the help of some people expert at PHP who found the solution to the problem. I had to modify the access rules of controller to give access to the required action that was being called in the URL parameter of Jquery $.post: array('allow', // allow authenticated user to perform 'dupecheck' actions 'actions'=>array('dupecheck'), 'users'=> array('authorised username'), That was a struggle!!

You could be right though because I have been struggling with this code and can't get it to work According to my Jquery reference book, $.post is supposed to do that. Waiting for some help from this forum

$.post initiates a POST request to the server...

Hi, I am having difficulty to have an instant validation carried out immediately after input of a value in a view file. Here is the code (extract of the view file) that calls for the function (found in the controller file) View file: <?php Yii::app()->getClientScript()->registerScript('checkid2','$("#thisid").change(function() { alert("actiondupecheck"); $.post("index.php?r=roomtype/dupecheck", {thisid: $("#thisid").val()}, function(data) { $("#dupecheck").html(data); }); } );'); ?> Controller file: public function actionDupecheck() { $thisid = trim($_POST['thisid']); $test =Piece::model()->find("id='$thisid'"); if(!empty($test)) { echo sprintf("This id <strong>%s</strong>, exists already.", $thisid); } } I have tried several ways but am completely stuck with this. I thought there was a problem with the syntax with the URL but after replacing 'dupecheck' with 'admin' in the view file, the admin function was called. What I am trying to do with this code, is to check for a duplicate entry in my table. I have been advised to use model validation rules. I will look into this suggestion but I also want to know that is wrong with my code. Thanks