使用Ajax发布到PHP并阅读结果

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);
}

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

echo json_encode($data);

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);
}

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);