Getting the path of fileupload/html input control

Hi ,
I have the following coding , which works well when i set path of the file manually.. i.e,
FileInfo fileInf = new FileInfo("C:\\new\\file1.zip");
  string uri = "ftp://" +
 ftpServerIP + "/" + fileInf.Name;
  FtpWebRequest reqFTP;
 
  // Create FtpWebRequest object from the Uri provided
  reqFTP =
 (FtpWebRequest)FtpWebRequest.Create(new Uri("ftp://" + ftpServerIP +
 "/" + fileInf.Name));
 
  // Provide the WebPermission Credintials
  reqFTP.Credentials = new NetworkCredential(ftpUserID, ftpPassword);
 
  // By default KeepAlive is true, where the control connection is not closed
  // after a command is executed.
  reqFTP.KeepAlive = false;
 
  // Specify the command to be executed.
  reqFTP.Method = WebRequestMethods.Ftp.UploadFile;
 
  // Specify the data transfer type.
  reqFTP.UseBinary = true;
 
  // Notify the server about the size of the uploaded file
  reqFTP.ContentLength = fileInf.Length;
 
  // The buffer size is set to 2kb
  int buffLength = 2048;
  byte[] buff = new byte[buffLength];
  int contentLen;
 
  // Opens a file stream (System.IO.FileStream) to read the file to be uploaded
  FileStream fs = fileInf.OpenRead();
 
  try
  {
  // Stream to which the file to be upload is written
  Stream strm = reqFTP.GetRequestStream();
 
  // Read from the file stream 2kb at a time
  contentLen = fs.Read(buff, 0, buffLength);
 
  // Till Stream content ends
  while (contentLen != 0)
  {
  // Write Content from the file stream to the FTP Upload Stream
  strm.Write(buff, 0, contentLen);
  contentLen = fs.Read(buff, 0, buffLength);
  }
 
  // Close the file stream and the Request Stream
  strm.Close();
  fs.Close();
  }
  catch (Exception ex)
  {
  MessageBox.Show(ex.Message, "Upload Error");
  }


in this coding , i set the path as fixed as ,

FileInfo fileInf = new FileInfo("C:\\new\\file1.zip");
but i want user to browse and select the file using fileupload or html input control. the problem is , i cant get the full path in fileupload. it just returns the filename alone and the above code returns the error "couldnot find the file". so pls help me how to achieve this by using fileupload control or anyother mreans.



Thanks,
Emm