How to return zip from bytearray on Servlet

Question

working again with Java EE.

In example

On servlet I have byte[] fileContent ---->(file.txt)
I try to add it to zip file

set servlet as

response.setContentType("Content-type: text/zip");
response.setHeader("Content-Disposition",
        "attachment; filename=mytest.zip");

// List of files to be downloaded
List files = new ArrayList();
files.add(new File("C:/first.txt"));

ServletOutputStream out = response.getOutputStream();
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(out));

for (File file : files) {

    System.out.println("Adding " + file.getName());
    zos.putNextEntry(new ZipEntry(file.getName()));

    // Get the file
    FileInputStream fis = null;
    try {
        fis = new FileInputStream(file);

    } catch (FileNotFoundException fnfe) {
        // If the file does not exists, write an error entry instead of
        // file
        // contents
        zos.write(("ERRORld not find file " + file.getName())
                .getBytes());
        zos.closeEntry();
        System.out.println("Couldfind file "
                + file.getAbsolutePath());
        continue;
    }

    BufferedInputStream fif = new BufferedInputStream(fis);

    // Write the contents of the file
    int data = 0;
    while ((data = fif.read()) != -1) {
        zos.write(data);
    }
    fif.close();

    zos.closeEntry();
    System.out.println("Finishedng file " + file.getName());
}

zos.close();

} }

I have problem with the second point. I try several of solutions but all of them hardly working with my servlet or didn't work.

Could somebody show me on example how add file as array of bytes to zip and return on servlet as atachment? Thanks for any tips and advices

--post edit

I try something like that, but how can I add byte[] instead of file.

Refer this link to ask good question http://stackoverflow.com/help/mcve — vels4j, Jun 02 '15 at 12:40

score 0 · Accepted Answer · answered Jun 03 '15 at 13:05

Here's example of solution:

public class ZipFileServlet extends HttpServlet {

private static final int DEFAULT_BUFFER_SIZE = 10240; // 10KB.
private YourFileDAO yourFileDAO = YourDAOFactory.getYourFileDAO();

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException
{
    String[] fileIds = request.getParameterValues("fileId");
    response.setContentType("application/zip");
    response.setHeader("Content-Disposition", "attachment; filename=\"allfiles.zip\"");
    ZipOutputStream output = null;
    byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];

    try {
        output = new ZipOutputStream(new BufferedOutputStream(response.getOutputStream(), DEFAULT_BUFFER_SIZE));

        for (String fileId : fileIds) {
            YourFileItem item = yourFileDAO.find(fileId);
            if (item == null) continue; // Handle yourself. The fileId may be wrong/spoofed.
            InputStream input = null;

            try {
                input = new BufferedInputStream(item.getInputStream(), DEFAULT_BUFFER_SIZE);
                output.putNextEntry(new ZipEntry(item.getName()));
                for (int length = 0; (length = input.read(buffer)) > 0;) {
                    output.write(buffer, 0, length);
                }
                output.closeEntry();
            } finally {
                if (input != null) try { input.close(); } catch (IOException logOrIgnore) { /**/ }
            }
        }
    } finally {
        if (output != null) try { output.close(); } catch (IOException logOrIgnore) { /**/ }
    }
}

 }

How to return zip from bytearray on Servlet

1 Answers1