Issue with using ajax to update without refresh - php

Question

I have this code here which is supposed to help me update a phone number.
It doesn't do it though, well, i get the successfully changed message but no insertion on the database.
Here is my code:
index.php
<script type="text/javascript" >
$(function() {
$(".submit").click(function() {
var phone = $("#phone").val();
var dataString = 'phone='+ phone ;
if(phone=='') {
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
} else {
$.ajax({
type: "POST",
url: "update-phone.php",
data: dataString,
success: function() {
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
});
</script>
<div class="modal" style="display: none;">
<?php
if (empty($phone)) {
?>
<form method="post" name="form">
<input id="phone" name="phone" type="text" />
<div>
<input type="submit" value="Submit" class="submit"/>
<span class="error" style="display:none"> Please Enter Valid Data</span>
<span class="success" style="display:none"> Registration Successfully</span>
</div>
</form>
update-phone.php
<?php
require_once('db.php');
if($_POST) {
$phone = $_POST['phone'];
mysql_query("UPDATE users SET phone = '$phone' WHERE ID = 5884 ");
}else {}
?>
What am i missing?
Thanks

Answer 1

Try this..
<?php
require_once('db.php');
if($_POST) {
$phone = $_POST['phone'];
mysql_query("UPDATE `users` SET `phone` = '$phone' WHERE `ID` = 5884 ");
}else {}
?>

Answer 2

Have you tried inspecting the ajax request with firebug/developer tools/etc?
Try adding echo mysql_error(); right after your mysql_query.
Not 100% sure but it could be the if ($_POST) should be replaced with if (isset($_POST['phone']))
Try the following php:
<?php
require_once('db.hp');
if (isset($_POST['phone']))
{
$phone = $_POST['phone'];
mysql_query("UPDATE users SET phone = '$phone' WHERE ID = 5884 ");
echo mysql_error();
}
else
{
echo "Failed";
}
?>
EDIT: Have you confirmed if it actually updates the DB?
Also you should also sanitise your input and consider rolling with mysqli instead of mysql_*