Question

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?

Answer 1

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) {
        }
    }
}