How can I get a resource "Folder" from inside my jar File?

Finally, I found the solution:

final String path = "sample/folder";final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());if(jarFile.isFile()) {  // Run with JAR file    final JarFile jar = new JarFile(jarFile);    final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar    while(entries.hasMoreElements()) {        final String name = entries.nextElement().getName();        if (name.startsWith(path + "/")) { //filter according to the path            System.out.println(name);        }    }    jar.close();} else { // Run with IDE    final URL url = Launcher.class.getResource("/" + path);    if (url != null) {        try {            final File apps = new File(url.toURI());            for (File app : apps.listFiles()) {                System.out.println(app);            }        } catch (URISyntaxException ex) {            // never happens        }    }}

The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.

Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:

URL url = MyClass.class.getResource("resources/");if (url == null) {     // error - missing folder} else {    File dir = new File(url.toURI());    for (File nextFile : dir.listFiles()) {        // Do something with nextFile    }}

You can also use

URL url = MyClass.class.getResource("/com/abc/package/resources/");

I know this is many years ago . But just for other people come across this topic.What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.