How do I create zip file in Servlet for download?

The example below is a servlet that shows you how to create a zip file and send the generated zip file for user to download. The compressing process is done by the zipFiles method of this class.

For a servlet to work you need to configure it in the web.xml file of your web application which can be found after the code snippet below.

package org.kodejava.example.servlet;

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 ZipDownloadServlet extends HttpServlet {
    public static final String FILE_SEPARATOR = System.getProperty("file.separator");

    protected void doPost(HttpServletRequest request, HttpServletResponse response) 
            throws ServletException, IOException {
        doGet(request, response);

    protected void doGet(HttpServletRequest request, HttpServletResponse response) 
            throws ServletException, IOException {
        try {
            // The path below is the root directory of data to be
            // compressed.
            String path = getServletContext().getRealPath("data");

            File directory = new File(path);
            String[] files = directory.list();

            // Checks to see if the directory contains some files.
            if (files != null && files.length > 0) {

                // Call the zipFiles method for creating a zip stream.
                byte[] zip = zipFiles(directory, files);

                // Sends the response back to the user / browser. The
                // content for zip file type is "application/zip". We
                // also set the content disposition as attachment for
                // the browser to show a dialog that will let user 
                // choose what action will he do to the sent content.
                ServletOutputStream sos = response.getOutputStream();
                response.setHeader("Content-Disposition", "attachment; filename="DATA.ZIP"");

        catch (Exception e) {

     * Compress the given directory with all its files.
    private byte[] zipFiles(File directory, String[] files) throws IOException {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ZipOutputStream zos = new ZipOutputStream(baos);
        byte bytes[] = new byte[2048];

        for (String fileName : files) {
            FileInputStream fis = new FileInputStream(directory.getPath() + 
                ZipDownloadServlet.FILE_SEPARATOR + fileName);
            BufferedInputStream bis = new BufferedInputStream(fis);

            zos.putNextEntry(new ZipEntry(fileName));

            int bytesRead;
            while ((bytesRead = != -1) {
                zos.write(bytes, 0, bytesRead);

        return baos.toByteArray();

The web.xml configuration:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="" 
         id="WebApp_ID" version="2.5">



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  1. Nice post, done exactly the same with my servlet.

    The files I need to ZIP are stored as BLOB and add them as byte[]. When I add them as a ZIPEntry everything works fine, but when I review the exported files, they don’t have a content type individually and marked as invalid content, any recommendations?

  2. Hi there,

    I used your code, it works great!, well I haven’t tested it with real data but it seems to be working fine, one question, where is the “data” folder in your example?

    Thank you!

  3. Hi Miguel,

    In this case the data directory should be under you webapp root directory. For example if your web application root directory is webapp then the data should be webapp/data. I hope that helps you.

  4. I tried running the code in jsp, then I can get a getOutputStream() has already been called for this response. any idea for this?

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