Loading Resources from a .JAR

Question

I've read through this forum and searched java.sun.com and have found nothing to solve my problem, although I see many other people are having the same problem.
I wrote a small application that needs to load a text file and images. When I just compile everything and then run it, all is well, my text file is found and my images are displayed. Now the problem is when I package everything into a .JAR file, suddenly my magic stops. Here is the directory structure
- State.class
FiftyStatesGame.class
FlagPanel.class
StatsPanel.class
StatePanel.class
states.txt
/images
/flags
a bunch of .gif images (50 total)
/states
a bunch of .gif images (50 total)
I create my archive with the following command
jar cvf FiftyStates.jar *
and all seems well, I've even changed the extension to .zip to ensure that the images and text file is inside the jar
Now when I try to run the program using
java -cp FiftyStates.jar FiftyStatesGame
the program starts but I get a NullPointerException and my pictures don't load. :-(
I've read in the forum about using
URL imageURL = ClassLoader.getSystemResource(<path to image>);
Image logo = Toolkit.getDefaultToolkit().getImage(imageURL);
and I tried this and once again, it works when not in jar form, but as soon as the application is JARed, I encounter the same problems.
Can anyone give me the answer to this crazy problem?
and I guess I should go one step further and ask, if I want to give this program to a friend with the JRE installed on a windows machine, what command should I put in my batch file to run this program on his machine?
javaw -cp FiftyStates.jar FiftyStatesGame
Thanks for any help and advice

Answer 1

Hi,
Getting the image
=================
It seems to me you can't be too far away from resolving your problem.
I've never actually tried to get an image from a jar file, so I'm guessing that this may not work for you, but I have unzipped a zip file contained within an executable jar.
What I did was this:
    private ZipInputStream findInstallationFile() {
        ClassLoader classLoader;
        URL url;
        ZipInputStream zipInputStream = null;
        InputStream inputStream;
        classLoader = getClass().getClassLoader();
        url = classLoader.getResource(ZIP_RESOURCE_NAME);
        if (url != null) {
            try {
                inputStream = url.openStream();
                zipInputStream = new ZipInputStream(inputStream);
            } catch (IOException ioe) {
                ioe.printStackTrace();
        } else {
            System.err.println("Failed to find resource: " + ZIP_RESOURCE_NAME);
        return zipInputStream;
    }The good thing about this is that you can see whether or not the resource was found in the jar file. If the returned URL is null, then the resource isn't there.
Once you have the InputStream, you can just read the entire contents and do what you like with them.
Executable jars
===============
You can make it easier to run a jar file by including a manifest file. This is just a text file which should be in the META-INF directory in your jar file. The manifest file itself should be called Manifest.mf.
The contents of the Manifest should be:
Manifest-Version: 1.0
Main-Class: FiftyStatesGame
This assumes that your main class is FiftyStatesGame which is in the default package (tut, tut!).
Hope this helps!
Rhys