How to check if a generated zip file is corrupted?

user_187309 0
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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 programatically 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?

zip java
8 Answers

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 (IOException e) {
        return false;
    } finally {
        try {
            if (zipfile != null) {
                zipfile.close();
                zipfile = null;
            }
        } catch (IOException e) {
        }
    }
}
user_2312890

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

user_2531250

in my implementation it looks like that. maybe it helps you:

//[...]

try {
    FileInputStream fis = new FileInputStream(file);
    BufferedInputStream bis = new BufferedInputStream(fis);

    zos.putNextEntry(new ZipEntry(file.getName()));

    try {
        final byte[] buf = new byte[BUFFER_SIZE];
        while (true) {
            final int len = bis.read(buf);
            if (len == -1) {
                break;
            }
            zos.write(buf, 0, len);
        }
        zos.flush();
        zos.closeEntry();
    } finally {
        try {
            bis.close();
        } catch (IOException e) {
            LOG.debug("Buffered Stream closing failed");
        } finally {
            fis.close();
        }
    }
} catch (IOException e) {
    throw new Exception(e);
}

//[...]
zos.close
user_5832100

Perhaps swap the following two lines?;

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

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

user_1971980

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.

user_2415900

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();
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