XMLHttpRequest post 传递多个参数及服务器端读取

一贯没搞定XMLHttpRequest  post方法怎样传递多种参数，比如一同读取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);      http://www.fpzhangsha.com     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']);    http://www.superkp123.com