How can I break up a large zip file into smaller zip files?

Question

I'd like to break up a single zip file that holds multiple files and create a zip file for each of the ZipEntries. That is: MainZipFile.zip ---> Zip1.zip, Zip2.zip, Zip3.zip Where the (right-hand-side) RHS zip files correspond to ZipEntrys inside MainZipFile.zip.

I'm using Java, and I'd like to do it without decompressing and re-compressing, which is the obvious solution. I'm not even sure if this is even practically possible.

I was thinking of creating empty RHS zip files, then using ZipInputStream to read the contents of an ZipEntry in MainZipFile.zip and deposit the contents into the corresponding RHS zip file, but I can't figure it out.

Edit 1:

final int BUFFER_SIZE_IN_BYTES=2*1024; //2KB
void tryReading(){
  String filepath = getTempFilepath();
  ZipFile mainZipFile = new ZipFile(filepath);
  InputStream inForEntry = getInputStreamForEntryWithName(checkpoint.getFilePath(), mainZipFile);
  ZipInputStream zin = new ZipInputStream(inForEntry);
  byte[] data = new byte[BUFFER_SIZE_IN_BYTES];
  int countOfBytesRead = zin.read(data); //returns -1
  zin.getNextEntry();
  countOfBytesRead = zin.read(data);//returns -1
}

private InputStream getInputStreamForEntryWithName(String name, ZipFile mainZipFile) throws IOException {
    ZipEntry entry = mainZipFile.getEntry(name);
    return mainZipFile.getInputStream(entry);
}

Answer 1

From the JavaDoc on the ZipInputStream read()

Reads from the current ZIP entry into an array of bytes. If len is not zero, the method blocks until some input is available; otherwise, no bytes are read and 0 is returned.

Returns: the actual number of bytes read, or -1 if the end of the entry is reached

I believe you're reading through the whole file and reaching the end. Wrap your read in a loop and read through a byte at a time and see if you can get it to output anything.

Answer 2

The InputStream you get from the ZipEntry gives you the result of decompressing the zip entry. Wrapping a ZipInputStream around it doesn't make any sense. The input the InputStream provides to the ZipInputStream isn't zipped data, it is plaintext. It can't possibly work.

In any case the data provided by ZipInputStream is more plaintext, not zipped data, so it can't possibly meet your objective.

You will have to read the InputStream and create a new zip file from it in the normal way you create zip files.

Answer 3

A better alternative seems to use a zip file system to "file copy" a selection of files from one to another zip.

This zip file system is already provided in Java SE for the jar: protocol. So if you have an URI file:/xxx/yyy.zip make an URI jar:file:/xxx/yyy.zip.

URI sourceZipURI = URI.create("jar:" + sourceZip); // "jar:file:/.../... .zip"
URI targetZipURI = URI.create("jar:" + targetZip); // 

Map<String, Object> senv = new HashMap<>(); 
FileSystem sourceZipFS = FileSystems.newFileSystem(sourceZipURI, senv, null);

Map<String, Object> tenv = new HashMap<>(); 
tenv.put("create", "true");
FileSystem targetZipFS = FileSystems.newFileSystem(targetZipURI, tenv, null);

// Example of a file copy:
Path sourcePath = sourceZipFS.getPath("/file1.txt");
Path targetPath = targetZipFS.getPath("/file1.txt"); // Or"/"
Files.copy(sourcePath, targetPath);

// Example of a zip traversal:
Files.walkFileTree(sourceZipFS.getPath("/"),
    new SimpleFileVisitor<Path>() {

        @Override
        public FileVisitResult visitFile(Path file,
                    BasicFileAttributes attrs)
                      throws IOException {
             ...
             return FileVisitResult.CONTINUE;
        }
    });

sourceZipFS.close();
targetZipFS.close();

A ZipInputStream/ZipOutputStream is possible too. For that search for some example code; as correct code would take some changes/additions.