Displaying xml content in the browser

Question

Hi,
In my application i am generating an XML content which i was storing into a file. Now as per the new requirement, this xml content needs to be displayed in the browser. (The way it gets displayed when we double click on an xml file).
Can some one plz tel me how to do this? I am using the jsp/servlet architechture..? plz help
Thanx

Answer 1

Program 1
import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class SendXml extends HttpServlet {
  public void doGet(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {
    //get the 'file' parameter
    String fileName = (String) request.getParameter("file");
    if (fileName == null || fileName.equals(""))
      throw new ServletException(
          "Invalid or non-existent file parameter in SendXml servlet.");
    // add the .doc suffix if it doesn't already exist
    if (fileName.indexOf(".xml") == -1)
      fileName = fileName + ".xml";
    String xmlDir = getServletContext().getInitParameter("xml-dir");
    if (xmlDir == null || xmlDir.equals(""))
      throw new ServletException(
          "Invalid or non-existent xmlDir context-param.");
    ServletOutputStream stream = null;
    BufferedInputStream buf = null;
    try {
      stream = response.getOutputStream();
      File xml = new File(xmlDir + "/" + fileName);
      response.setContentType("text/xml");
      response.addHeader("Content-Disposition", "attachment; filename="
          + fileName);
      response.setContentLength((int) xml.length());
      FileInputStream input = new FileInputStream(xml);
      buf = new BufferedInputStream(input);
      int readBytes = 0;
      while ((readBytes = buf.read()) != -1)
        stream.write(readBytes);
    } catch (IOException ioe) {
      throw new ServletException(ioe.getMessage());
    } finally {
      if (stream != null)
        stream.close();
      if (buf != null)
        buf.close();
  public void doPost(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {
    doGet(request, response);
Program 2
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class ResourceServlet extends HttpServlet {
  public void doGet(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {
    //get web.xml for display by a servlet
    String file = "/WEB-INF/web.xml";
    URL url = null;
    URLConnection urlConn = null;
    PrintWriter out = null;
    BufferedInputStream buf = null;
    try {
      out = response.getWriter();
      url = getServletContext().getResource(file);
      //set response header
      response.setContentType("text/xml");
      urlConn = url.openConnection();
      //establish connection with URL presenting web.xml
      urlConn.connect();
      buf = new BufferedInputStream(urlConn.getInputStream());
      int readBytes = 0;
      while ((readBytes = buf.read()) != -1)
        out.write(readBytes);
    } catch (MalformedURLException mue) {
      throw new ServletException(mue.getMessage());
    } catch (IOException ioe) {
      throw new ServletException(ioe.getMessage());
    } finally {
      if (out != null)
        out.close();
      if (buf != null)
        buf.close();
  public void doPost(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {
    doGet(request, response);
}Have fun