How to get a path to a resource in a Java JAR file

This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.

Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.

When loading a resource make sure you notice the difference between:

getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path

and

getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning

I guess, this confusion is causing most of problems when loading a resource.

Also, when you're loading an image it's easier to use getResourceAsStream():

BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));

When you really have to load a (non-image) file from a JAR archive, you might try this:

File file = null;String resource = "/com/myorg/foo.xml";URL res = getClass().getResource(resource);if (res.getProtocol().equals("jar")) {    try {        InputStream input = getClass().getResourceAsStream(resource);        file = File.createTempFile("tempfile", ".tmp");        OutputStream out = new FileOutputStream(file);        int read;        byte[] bytes = new byte[1024];        while ((read = input.read(bytes)) != -1) {            out.write(bytes, 0, read);        }        out.close();        file.deleteOnExit();    } catch (IOException ex) {        Exceptions.printStackTrace(ex);    }} else {    //this will probably work in your IDE, but not from a JAR    file = new File(res.getFile());}if (file != null && !file.exists()) {    throw new RuntimeException("Error: File " + file + " not found!");}

The one line answer is -

String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()

Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().