Question

I am trying to zip a group of binary data (result set returned from database) into a single file. Which can be downloaded via web application. Following code is used to zip the result set and write the zip file to HttpServletResponse

String outFilename = "outfile.zip";
response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition", "attachment; filename= " + outFilename);
OutputStream os = response.getOutputStream();
ZipOutputStream out = new ZipOutputStream(os);
for (int i = 0; i < cardFileList.size(); i++) {

    CardFile cardFile = cardFileList.get(i);
    out.putNextEntry(new ZipEntry(cardFile.getBinaryFileName()));
    out.write(cardFile.getBinaryFile(), 0, cardFile.getBinaryFile().length);
    out.closeEntry();
}
// Complete the ZIP file
out.flush();
out.close();
os.close();

The problem is that while unzipping the downloaded zip file using WinRar I get following error :

File Path: Either multipart or corrupt ZIP archive

Can someone point out where am I making mistake?. Any help would be appreciated.

[EDIT] I tried response.setContentType("application/zip"); but same result.

Answer 1

The following code works for me:

FileOutputStream os = new FileOutputStream("out.zip");
ZipOutputStream zos = new ZipOutputStream(os);
try
{
    for (int i = 1; i <= 5; i++)
    {
        ZipEntry curEntry = new ZipEntry("file" + i + ".txt");
        zos.putNextEntry(curEntry);
        zos.write(("Good morning " + i).getBytes("UTF-8"));
    }
}
finally
{
    zos.close();
}

Zip files generated with this code opens up with 7-zip without problem.

• Check that the response from the servlet is not actually a 404 or 500 error page. Pick a small response and open it up with a hex editor or even a text editor. Zip files start with a ‘PK’ magic number which should be visible even in a text editor.
• Try saving to file instead of servlet output stream for starters and see if that makes a difference.
• Could there be a filter modifying your servlet’s output / corrupting the ZIP?