ZipEntry to File

Question

Is there a direct way to unpack a java.util.zip.ZipEntry to a File?

I want to specify a location (like "C:\temp\myfile.java") and unpack the Entry to that location.

There is some code with streams on the net, but I would prefer a tested library function.

Solution 1

Use ZipFile class

    ZipFile zf = new ZipFile("zipfile");

Get entry

    ZipEntry e = zf.getEntry("name");

Get inpustream

    InputStream is = zf.getInputStream(e);

Save bytes

    Files.copy(is, Paths.get("C:\\temp\\myfile.java"));

Solution 2

Use the below code to extract the "zip file" into File's then added in the list using ZipEntry. Hopefully, this will help you.

private List<File> unzip(Resource resource) {
    List<File> files = new ArrayList<>();
    try {
        ZipInputStream zin = new ZipInputStream(resource.getInputStream());
        ZipEntry entry = null;
        while((entry = zin.getNextEntry()) != null) {
            File file = new File(entry.getName());
            FileOutputStream  os = new FileOutputStream(file);
            for (int c = zin.read(); c != -1; c = zin.read()) {
                os.write(c);
            }
            os.close();
            files.add(file);
        }
    } catch (IOException e) {
        log.error("Error while extract the zip: "+e);
    }
    return files;
}

Solution 3

Use ZipInputStream to move to the desired ZipEntry by iterating using the getNextEntry() method. Then use the ZipInputStream.read(...) method to read the bytes for the current ZipEntry. Output those bytes to a FileOutputStream pointing to a file of your choice.

Comments

[J Fabian Meier]

This means: No, there is no library function?

[Acuna]

Yes, @JFMeier yes, right you) There's much of library functions which Java don't have.