获取null`request.parts`通过jQueryAjax将文件发布到Javaservlet

获取null`request.parts`通过jQueryAjax将文件发布到Javaservlet,java,jquery,ajax,servlets,Java,Jquery,Ajax,Servlets,以下是我的js代码: function sendMP3ViaAJAX(MP3File) { console.log(MP3File.name); //file is present, as I can get its name $.ajax({type: "POST", url: "saveMP3", enctype: 'multipart/form-data', processData: false, data

以下是我的js代码:

function sendMP3ViaAJAX(MP3File) {
    console.log(MP3File.name); //file is present, as I can get its name
      $.ajax({type: "POST",
        url: "saveMP3",
        enctype: 'multipart/form-data',
        processData: false,
        data: {'mp3File': MP3File},
        success: function (text)
        {
            response = text;
        },
        complete: function () {
             $("#retourAJAX").html("blah blah");
        }
    });
}
但是如果我调试我的项目

 @Override
    public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {


        Part part = request.getPart("MP3FILE"); //request.parts is null...


(常数
CHAMP_FICHIER
等于AJAX调用中定义的值,即
mp3File
..)您知道缺少什么吗?由于我已经在页面上动态加载了内容,我不希望出现默认的HTML POST submit行为…

您需要以二进制数据而不是编码字符串的形式发送文件。看看你是如何使用
FormData
上传的,谢谢你的帮助!