我试图通过Ajax从服务器检索XML文件，但无法将其解析，我不知道是哪出了问题？当我调用getFriends.php时，它会打印XML。以下是Ajax代码：
<!DOCTYPE>
<html>
<head>
<title>Ajax Sample</title>
<script type="text/javascript">
function getFriendsList(MemberId) {
var xmlhttp;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else {
// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
var xmlDoc = xmlhttp.responseXML;
var friendElements = xmlDoc.getElementsByTagName('friend');
for (var x=0; x<friendElements.length; x++) {
// We know that the first child of show is title, and the second is rating
var id = showElements[x].childNodes[0].value;
var name = showElements[x].childNodes[1].value;
// Now do whatever you want with the show title and ratings
document.write("hi");
}
}
}
xmlhttp.open("POST","getFriends.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("id=" + MemberId);
}
</script>
</head>
<body>
<input name="ajax" type="button" onClick="getFriendsList(1)" value="Click for AJAX">
<div style="background-color:#00FF99;" id="placehere">Here is where the update will occur.</div>
</body>
</html>
这是getFriends.php代码（用户类运行正常）：
<?php
include('lib.php');
//$id = $_POST['id'];
$id=1;
$user = new User($id);
echo $user->getFriendsList();
?>
