Problem with session while sending a multipart-form (a file via post)

Question

I have a problem with the communication between my applet and my server. I use a method to post an xml to my server with the Content-Type "application/x-www-form-urlencoded" and with it I'm not having any problem to preserve the session.
However I use another method I found in the forums to post files and try to adapt it, but I'm being send from the server to the login page because it's not recognizing my session
Can anybody help me? This is my method
public String uploadFile(String adr, BufferedImage img) {
        String res = "";
        String CONTENT_BOUNDARY = "--abvx98734732";
        String s;
        File f = null;
        URL url;
        URLConnection conn;
        try {
            url = new URL(adr);
            conn = url.openConnection();
            conn.setDoOutput(true);
            conn.setRequestProperty("Content-Type",
                    "multipart/form-data; boundary=" + CONTENT_BOUNDARY);
            conn.setRequestProperty("file-Name", "process_image.jpg");
            DataOutputStream wr = new DataOutputStream(conn.getOutputStream());
            wr
                    .write(("--"
                            + CONTENT_BOUNDARY
                            + "\r\n"
                            + "Content-Disposition: form-data; name=\"fileData\"; filename=\""
                            + "process_image.jpg" + "\"\r\nContent-Type: image/jpg\r\n\r\n")
                            .getBytes());
            ImageIO.write(img, "jpg", wr);
            wr.write(("\r\n--" + CONTENT_BOUNDARY + "--\r\n").getBytes());
            wr.flush();
            BufferedReader in = new BufferedReader(new InputStreamReader(conn
                    .getInputStream()));
            while ((s = in.readLine()) != null) {
                res += s;
            in.close();
        catch (MalformedURLException muo) {
            muo.printStackTrace();
        catch (IOException io) {
            io.printStackTrace();
        return res;
    }This is how I preserve session in my other method:
private static final String SESSION_KEY = "_session_id";
URLEncoder.encode(SESSION_KEY, "UTF-8") + "=" +
URLEncoder.encode(sessionId, "UTF-8");
wr.writeBytes(content);
....

Answer 1

Yes, the only way is to pass the value u want in the query string. But make sure that the value is not too large.
In case you have too much info i guess the only way is to circumvent multipart and handle the process in your own custom class, which wont be worth the effort if u need pass only 2-3 small parameters.
Always,
Leo.
Hi,
Before I upload a file, I want to get a few parameter
values (getParameter) on the form first. I've tried
using com.oreilly.servlet MultipartRequest and a few
other free servlets but they require that the file
must to be uploaded first. For example:
MultipartRequest multi = new
MultipartRequest(request,filepath, 10*1024*1024);
//this will instantiate the object and upload the
file
String command = multi.getParameter("command"); //get
the form parameter
Is there any way I can get the parameter first before
uploading the file? I truly appreciate any helps you
can provide me.
Regards,
Lily