Question

I have a method which zips up 5 files. It produces a zip file without error, but I cannot open it to examine the contents. I tried emailing it and gmail said it cannot send corrupt files. Trying to open with WinRAR in Windows results in an error stating:

The archive is either in unknown format or damaged

This is the method:

private void zipTestFiles() throws FileNotFoundException, IOException
{
    File[] filenames = fileDir.listFiles(fileNameFilter(Constants.PAGE_MON_FILENAME_FILTER));

    byte[] buf = new byte[1024];

    String outFilename = Environment.getExternalStorageDirectory().getAbsolutePath() + File.separatorChar + Constants.PAGEMONITOR_ZIP;
    DeflaterOutputStream out = new DeflaterOutputStream(new BufferedOutputStream(new FileOutputStream(outFilename)));

    for (int i=0; i<filenames.length; i++)
    {
        FileInputStream in = new FileInputStream(filenames[i]);

        int len;
        while ((len = in.read(buf)) > 0)
        {
            out.write(buf, 0, len);
        }

        in.close();
        }
    out.close();
}

Answer 1

You should use ZipOutputStream instead of DeflaterOutputStream. And do not forget to create entries. Read javadoc of ZipOutputStream before writing the implementation.