Recipe 13.3 Sending an XML file

Problem

You want to send an XML file as binary data from a servlet.

Solution

Use the javax.servlet.ServletOutputStream obtained from the javax.servlet.http.HttpServletResponse object to send the XML file as binary data to the client.

Discussion

This recipe describes how to send an XML file as binary data from a ServletOutputStream, so that the user can handle the file as downloaded XML. Example 13-3 obtains the bytes that represent the XML as a BufferedInputStream wrapped around a FileInputStream. The code is very similar to Example 13-1 in Recipe 13.1, except that it uses a MIME type of text/XML.

In a popular form of converting XML into a readable format, you could convert the XML content to HTML or another form using Extensible Stylesheet Language Transformations (XSLT). If the intent is to use XSLT for generating the content in a browser, leave out the Content-Disposition response header, because this header is designed to handle the XML as a downloaded file that will be saved in the user's filesystem. See Chapter 23 on using the x:transform JSTL tag.

Example 13-3. Sending an XML file with a servlet

package com.jspservletcookbook;           

import java.io.FileInputStream;
import java.io.BufferedInputStream;
import java.io.File;
import java.io.IOException;

import javax.servlet.*;
import javax.servlet.http.*;

public class SendXml extends HttpServlet {

  public void doGet(HttpServletRequest request, 
    HttpServletResponse response) throws ServletException, 
      IOException {
    
     //get the filename from 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 .xml suffix if it doesn't already exist
     if (fileName.indexOf(".xml") == -1)
          fileName = fileName + ".xml";
          
     //where are XML files kept?
     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);
     
         //set response headers
         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;

         //read from the file; write to the ServletOutputStream
         while((readBytes = buf.read( )) != -1)
            stream.write(readBytes);

     } catch (IOException ioe){
     
        throw new ServletException(ioe.getMessage( ));
         
     } finally {
     
         //close the input/output streams
         if(stream != null)
             stream.close( );

         if(buf != null)
             buf.close( );

      }//finally
    
  } //end doGet
   
  public void doPost(HttpServletRequest request, 
    HttpServletResponse response)
      throws ServletException, IOException {
        
        doGet(request,response);
  } 
}

For the context-param to work correctly in this code, you have to include in web.xml an element that looks like:

<context-param>
    <param-name>xml-dir</param-name>
    <param-value>h:/home/xml</param-value>
</context-param>

See Chapter 1 if you need an introduction or refresher for the deployment descriptor web.xml.

The discussion in Recipe 13.1 describes the basic mechanics of this code, so I don't repeat that information here. See the note at the end of Recipe 13.1 about the Internet Explorer-related exception that you may experience with servlets of this type.

Recipe 13.3 Sending an XML file

Problem

Solution

Discussion

Example 13-3. Sending an XML file with a servlet

See Also