ZipEntry 到文件

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.

Answer 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"));

Answer 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;
}

Answer 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.