XMLHttpRequest post 传递多个参数及服务器端读取

一直没搞定XMLHttpRequest post方法如何传递多种参数，比如同时读取post参数和file参数

var http = new XMLHttpRequest();

var form = new FormData();

// Add selected file to form

form.append(me.getName(), file);

form.append('filename', '1.png');

// Send form with file using XMLHttpRequest POST request

http.open('POST', me.getUrl());

http.send(form);

服务器端

$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);

$uploadfile = $uploaddir . basename($_POST['filename']);