I am Student and having a huge problem need some help quick!!!

Question

Hello I am student in college and i am developing a file share portal for my final year project using Flex 3 on the client side, java on the server side, hibernate for ORM, Sping MVC and My Sql 5. I am having a bit of a problem parsing this:
------------GI3ae0ae0ae0GI3KM7KM7Ef1cH2ei4
Content-Disposition: form-data; name="Filename"
Flex3.txt
------------GI3ae0ae0ae0GI3KM7KM7Ef1cH2ei4
Content-Disposition: form-data; name="action"
upload
------------GI3ae0ae0ae0GI3KM7KM7Ef1cH2ei4
Content-Disposition: form-data; name="file"; filename="Flex3.txt"
Content-Type: application/octet-stream
Flex 3 can be a pain in the arse!!!!!!
------------GI3ae0ae0ae0GI3KM7KM7Ef1cH2ei4
Content-Disposition: form-data; name="Upload"
Submit Query
------------GI3ae0ae0ae0GI3KM7KM7Ef1cH2ei4--
This is my Java code:
package com.file.exchange.service;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.HttpSession;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import java.io.*;
import java.sql.Blob;
import com.file.exchange.vo.Files;
import java.util.Date;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.File;
import org.apache.commons.fileupload.*;
import org.apache.commons.fileupload.disk.*;
import org.apache.commons.io.*;
import org.apache.commons.fileupload.servlet.*;
import java.util.Iterator;
import java.util.List;
import java.util.ArrayList;
import java.util.ListIterator;
import javax.imageio.ImageIO;
import javax.imageio.ImageReader;
import javax.imageio.stream.ImageInputStream;
import org.apache.commons.fileupload.util.FileItemHeadersImpl;
import javax.servlet.*;
import javax.servlet.http.*;
public class FileUpload extends FileBaseService
     InputStream in = null;
     OutputStream out = null;
     ByteArrayOutputStream bout = null;
     @SuppressWarnings("unchecked")
     @Override
     protected String getXmlFromOperation(HttpServletRequest request,HttpServletResponse response) throws Exception
          int readBytes = -1;
int length = request.getContentLength();
in = request.getInputStream();
byte[] buffer = new byte[length];
bout = new ByteArrayOutputStream(length);
while((readBytes = in.read(buffer, 0, length)) != -1) {
bout.write(buffer, 0, readBytes);
byte[] inData = bout.toByteArray();
FileOutputStream fos = new FileOutputStream("C:/Documents and Settings/Andrew Hobbs/Desktop/helloworld.txt");
for(int i=0; i<inData.length; i++)
fos.write(inData);
fos.close();
          DateFormat dateFormat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
          Date date = new Date();
          String data = "Hello";
          String id = "1";//getCharacterDataFromElement(vID);
          String name = "meh";//getCharacterDataFromElement(vname);
          String type = "1";//getCharacterDataFromElement(vtype);
          String size = "5555";//Long.toString(fileObject.length());//getCharacterDataFromElement(vsize);
          Blob content = org.hibernate.Hibernate.createBlob(data.getBytes());//fisFile);//data.getbytes[]
          String date_uploaded = dateFormat.format(date);
          Files file = getFileDao().uploadFile(Integer.parseInt(id), name, type, Integer.parseInt(size), content, date_uploaded);
          return file.toXml();
I will so grateful if somebody could point me in the right direction with this or point me to possible examples of parsing this!
Thanks

Answer 1

Let me warn you up front: never mark your questions as urgent. People really don't care how urgent it is to you, and will likely only irritate them enough to not want to help you. Lucky for you, I couldn't care less ;)
you seem to want to get an uploaded file using FileUpload. Did you see the user guide on how to do this? You seem to be going about it in completely the wrong way.
[FileUpload user guide|http://commons.apache.org/fileupload/using.html]
Also:
FileOutputStream fos = new FileOutputStream("C:/Documents and Settings/Andrew Hobbs/Desktop/helloworld.txt");Hardcoded paths in your code? Not very portable is it. I would define paths in a properties file. Read this:
[smartly loading properties|http://www.javaworld.com/javaworld/javaqa/2003-08/01-qa-0808-property.html]