Jquery to get php mysql results

Jquery to get php mysql results

SO I have successfully gotten my form to submit but I need to display my comments with each image in a gallery based on the image name attribute which is stored in a db and passed to jquery via javascript variable.
I cannot seem to get the comments to display in the browser and I get no error messages.
here is my jquery code:
      var curRow = dsCottonClubMain.getCurrentRow();
      var curRowNum = curRow['@id'];
      var curName = curRow['name'];
       var queryString = 'index.php?name=' + curName;
      //alert('Image Name is ' + curName);
   
function showCmnts (){
   $.ajax({
   
   type: "POST",
   url: "showCommentsPHP.php",
   data: queryString,
   success: function(data){
      $("#showComments").showResponse(responseText, statusText)
      },
   
   error: function(){
      $("#showComments").p("An Error Occurred");}
      });
};

MY JQUERY CODE
function getCmnts() {
    var options = {
        target:        '#showComments',
        beforeSubmit:  showRequest,
        success:       showResponse,
        type:          'post'
    };
    // bind to the form's submit event
    $('#getCommentsForm').submit(function() {
        // inside event callbacks 'this' is the DOM element so we first
        // wrap it in a jQuery object and then invoke ajaxSubmit
        $(this).ajaxSubmit(options);

        // always return false to prevent standard browser submit and page navigation
        return false;
    });
};

//// CALLBACKS -- // ---->


// pre-submit callback
function showRequest(formData, jqForm, options) {
    // formData is an array; here we use $.param to convert it to a string to display it
    // but the form plugin does this for you automatically when it submits the data
    var queryString = $.param(formData); 
    // jqForm is a jQuery object encapsulating the form element.  To access the
    // DOM element for the form do this:
    // var formElement = jqForm[0]; 
    return true;
}


// post-submit callback
function showResponse(responseText, statusText)  {
    // for normal html responses, the first argument to the success callback
    // is the XMLHttpRequest object's responseText property
    // if the ajaxSubmit method was passed an Options Object with the dataType
    // property set to 'xml' then the first argument to the success callback
    // is the XMLHttpRequest object's responseXML property 
}

here is my php:
$cmnt = mysql_query("SELECT imgName,usrName,usrEmail,usrComment,date
                  FROM comments
                  WHERE imgName =" . $_POST['name'];);
   
   while($data = mysql_fetch_array($cmnt)
   {
   $img = $data[0];
   $uName = $data[1];
   $uEmail = $data[2];
   $uComment = $dtat[3];
   $uDate = $data[4];

echo "<div>
      <ul>
        <li>Posted: " . $uDate ."</li>
        <li>By: ". $uName ."</li>
     </ul>
     <p> ". $uComment  ."</p>
</div>"


   }
?>

Not sure what I have done here but it just doesn't give me anything back from the server.

Tahnx