处理多部分/表单数据

Question

I'm doing file upload using XMLHttpRequest() in my jsp and when I do request.getContentType() in my controller I'm getting:

multipart/form-data; boundary=---------------------------4664151417711.

Further I'm not getting how to get the file and get the contents of it in my controller. Please anyone help.

Update -- I'm doing this in my jsp.

 function fileUpload() {
var url= document.getElementById("urlId").value;
 var file= document.getElementById("xslId").files[0];
 var formdata = new FormData();
 formdata.append("url", url);
 formdata.append("file", file);
 var xhr = new XMLHttpRequest();       
 xhr.open("POST","http://localhost:8080/XlsUpload/openSource.htm", true);
 xhr.send(formdata);
 xhr.onload = function(e) {
    };                    
   }   

and in my controller--

 public void openSource(@ModelAttribute("domTool") DomTool   domTool,HttpServletRequest     request,HttpServletResponse response){
     String type=request.getContentType();

Further I'm struck how to get the contents of the uploaded file and the value of text field ie,URL in my controller. The type i'm getting as multipart/form-data

Answer 1

There is an Apache commons solution called commons-fileupload for parsing multipart content. You can find it here .

The most simple example copied from their tutorial looks like this:

@Override
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    FileItemFactory factory = new DiskFileItemFactory();
    ServletFileUpload upload = new ServletFileUpload(factory);
    List items = upload.parseRequest(request);

    // iterate over items (i.e. list of FileItem) and access 
    // the content with getInputStream()
}