上传文件jQuery 的ajaxFileUpload 方法上传文件

需要的js:
    <script src="jquery-1.7.1.js" type="text/javascript"></script>

    <script src="ajaxfileupload.js" type="text/javascript"></script>

jsp:
   <input type="file" id="filetest" name="filetest" title="我是input的file框"/>
  <input type="button" id="button" value="上传测试" />

js:
   $(document).ready(function(){
$("#button").click(function(){
ajaxFileUpload();
});
})

function ajaxFileUpload(){
alert("金乌");
$.ajaxFileUpload({
url:"${pageContext.request.contextPath}/vulnprotectinfo_uploadTest.do",
fileElementId:"filetest",
data:{"data":"23"},
type:"post",
success:function(data){
alert("上传成功!");
alert(data.rs);
}
});
}

action:








        private File filetest;          //id的名字
private String filetestFileName;//文件的名字,最好这样命名,不这样命名也可以得到这个数据
private String filetestContentType;//文件类型


public void uploadTest(){
HttpServletRequest request = ServletActionContext.getRequest();
HttpServletResponse response = ServletActionContext.getResponse();
String data = request.getParameter("data");
String rs="false";
String path = ServletActionContext.getServletContext().getRealPath("updateLoadTest");
File filePath = new File(path);
if(!filePath.exists()){
filePath.mkdirs();
}
// System.out.println(filetestFileName);
String uploadPath=filePath+File.separator+filetestFileName;
File upload=new File(uploadPath);
InputStream in=null;
OutputStream out = null;
PrintWriter pw=null;
try {
pw=response.getWriter();
in=new BufferedInputStream(new FileInputStream(filetest),6 * 1024);
out=new BufferedOutputStream(new FileOutputStream(upload), 6*1024);
byte[] buff=new byte[6* 1024];
while (in.read(buff)>0) {
out.write(buff);
}
rs="true";
pw.print(rs);
} catch (Exception e) {
pw.print(rs);
e.printStackTrace();
}finally{
try {
in.close();
out.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}