$.get PHP file in JS when link clicked - php

Question

The link:
<a class='dropdown' href='javascript:void(0);' onclick='mainLoginToggle();'>
mainLoginToggle():
function mainLoginToggle() {
$(document).mouseup(function (e) {
var container = $(".logindisplay");
if (container.has(e.target).length === 0) {
container.fadeOut(222);
}
});
if (document.getElementById('logindisplay').style.display == 'block') {
$(".logindisplay").fadeOut(222);
} else {
$(".logindisplay").fadeIn(222);
}
$.get("../include/removenotif.php");
return false;
}
removenotif.php:
<?php include("session.php"); $database->removeNotifications($session->username); ?>
removeNotification():
function removeNotifications($user) {
$q = "UPDATE notifications SET seen = '1' WHERE username = '$user'";
$result = mysql_query($q, $this->connection);
return true;
}
Basically, when the link is clicked it shows and hides the div, but also when it's clicked it's supposed to run the removeNotifications function, and not return any messages but successfully run the sql query. But it's not executing the query.
I've tested to see if the page is actually being called by changing the header information to try and get an error but nothing is happening. I'm not really sure where I'm going wrong here, any help would be appreciated. Thanks.

Answer 1

Why not use JQuery all the way?
Like this
Your link
<a class="dropdown" href="#">Click here</a>
Jquery
$(document).ready(function() {
$('.dropdown').live('click', function(){
var container = $(".logindisplay");
if (container.has(e.target).length === 0) {
container.fadeOut(222);
}
if (document.getElementById('logindisplay').style.display == 'block') {
$(".logindisplay").fadeOut(222);
} else {
$(".logindisplay").fadeIn(222);
}
$.get("../include/removenotif.php");
return false;
});
});
This way the PHP file is called.
Wezy