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.

*"...to a `File`?"* [`File`](http://docs.oracle.com/javase/8/docs/api/java/io/File.html) is an abstract representation of a pathname. Do you mean "file" (no code markup)? — T.J. Crowder, Apr 27 '16 at 12:40
A mean, I give the function a File and it unpacks the entry to that abstract pathname. — J Fabian Meier, Apr 27 '16 at 12:41
I am sorry if this is unclear or a duplicate, but please comment instead of justing voting it down. — J Fabian Meier, Apr 27 '16 at 12:43

score 29 · Accepted Answer · answered Apr 27 '16 at 12:54

29

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"));

answered Apr 27 '16 at 12:54

Evgeniy Dorofeev

133,369
30
199
275

Sathiamoorthy · Answer 2 · 2023-04-03T14:02:02.260

Use the below code to extract the "zip file" into File's then add 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;
}

score 1 · Answer 3 · answered Apr 27 '16 at 12:48

1

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.

answered Apr 27 '16 at 12:48

Coffee Monkey

481
2
7

This means: No, there is no library function? – J Fabian Meier Apr 27 '16 at 12:51
Yes, @JFMeier yes, right you) There's much of library functions which Java don't have. – Acuna Jul 19 '18 at 19:15

ZipEntry to File

3 Answers3

Linked