Post to PHP with Ajax and read Results - php

Question

I have read through dozens of similar questions on this website, and am having a lot of trouble trying to understand what is wrong with my code. I am trying to dynamically update select boxes based on the value of another one, but for some reason cannot seem to get any type of response data back once I post to PHP with Ajax.
JAVASCRIPT:
function toggleHiddenDIV()
{
var dc = document.getElementById("datacenter");
var dcVal = dc.options[dc.selectedIndex].value;
// Check if Datacenter selection has no Value selected
if(dcVal != '')
{
document.getElementById("hidden-options").style.display="block";
$.ajax({
type: "POST",
url: "handler.php",
data: { 'action_type': 'update_inventory_fields', id: dcVal },
success: function(response)
{
alert(response);
}
});
}
else
{
document.getElementById("hidden-options").style.display="none";
}
};
</script>
PHP:
if ($_POST['action_type'] == "update_inventory_fields")
{
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["id"])) { return; }
}
$result = mysql_query("SELECT id, ip, block FROM ipv4 WHERE datacenter = " . $_POST["id"]);
$data = array();
while($row = mysql_fetch_array($result, true))
{
$data[] = $row;
};
return json_encode($data);
}

Answer 1

Don't call return (since you're not returning a function); just echo then content onto the page:
echo json_encode($data);

Answer 2

Change to this...no need to return, just echo, since youre outside of a function call
if ($_POST['action_type'] == "update_inventory_fields")
{
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["id"])) { return; }
}
$result = mysql_query("SELECT id, ip, block FROM ipv4 WHERE datacenter = " . $_POST["id"]);
$data = array();
while($row = mysql_fetch_array($result, true))
{
$data[] = $row;
};
echo json_encode($data);
}

Answer 3

If the php code you posted is inside a function than you need to use echo functionname();
If the php code is not in a function, then just use echo json_encode($data);