Java: how to check if a generated zip file is corrupted?

Question

Hi,

we have a piece of code which generates a zip file on our system. Everything is ok, but sometimes this zip file while opened by FilZip or WinZip is considered to be corrupted.

So here is my question: how can we check programmatically if a generated zip file is corrupted?

Here is the code we are using to generate our zip files:

try {
                ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(tmpFile));
                byte[] buffer = new byte[16384];
                int contador = -1;
                for (DigitalFile digitalFile : document.getDigitalFiles().getContent()) {
                    ZipEntry entry = new ZipEntry(digitalFile.getName());
                    FileInputStream fis = new FileInputStream(digitalFile.getFile());
                    try {
                        zos.putNextEntry(entry);
                        while ((counter = fis.read(buffer)) != -1) {
                            zos.write(buffer, 0, counter);
                        }
                        fis.close();
                        zos.closeEntry();
                    } catch (IOException ex) {
                        throw new OurException("It was not possible to read this file " + arquivo.getId());
                    }
                }
                try {
                    zos.close();
                } catch (IOException ex) {
                    throw new OurException("We couldn't close this stream", ex);
                }

Is there anything we are doing wrong here?

EDIT: Actually, the code above is absolutely ok. My problem was that I was redirecting the WRONG stream for my users. So, instead of opening a zip file they where opening something completely different. Mea culpa :(

BUT the main question remains: how programatically I can verify if a given zip file is not corrupted?

Answer 1

I think you'll see correspondent exception stack trace during zip-file generation. So, you probably wan't to enhance your exception handling.

Answer 2

ZipOutputStream does not close the underlying stream.

What you need to do is:

FileOutputStream fos = new FileOutputStream(...);
ZipOutputStream zos = new ZipOutputStream(fos);

Then in your closing block:

zos.close();
fos.flush(); // Can't remember whether this is necessary off the top of my head!
fos.close();

Answer 3

Perhaps swap the following two lines?;

fis.close();
zos.closeEntry();

I can imagine that the closeEntry() will still read some data from the stream.

Answer 4

Your code is basically OK, try to find out which file is responsible for the corrupted zip file. Check whether digitalFile.getFile() always returns a valid and accessible argument to FileInputStream. Just add a bit logging to your code and you will find out what's wrong.

Answer 5

You can use the ZipFile class to check your file :

 static boolean isValid(final File file) {
    ZipFile zipfile = null;
    try {
        zipfile = new ZipFile(file);
        return true;
    } catch (ZipException e) {
        return false;
    } catch (IOException e) {
        return false;
    } finally {
        try {
            if (zipfile != null) {
                zipfile.close();
                zipfile = null;
            }
        } catch (IOException e) {
        }
    }
}