更新数据库后的响应页面

Question

如何使以下代码被AJAX调用，即，单击链接以更新数据库，并在浏览器中更新页面而不重新加载它？

<?php

$server = "localhost";
$user = "root";
$pass = "";
$database = "test";

mysql_connect($server, $user, $pass) or die ("check mysql settings");
mysql_select_db($database) or die ("database not found");

$data = mysql_query("SELECT * FROM settings ");
while ($row = mysql_fetch_array($data)) {
    $images = $row["images"];
    if($images == "yes") {
        echo "<a href='index.php?img=no'>Do not show images on this site</a>";
    } else {
        echo "<a href='index.php?img=yes'>Show images on this site</a>";
    }
}

$img = $_GET["img"];
if ($img == "yes") {
    mysql_query("update settings set images='yes'");
} else {
    mysql_query("update settings set images='no'");
}