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']);