获取null`request.parts`通过jQueryAjax将文件发布到Javaservlet
以下是我的js代码: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
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
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