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

Question

I am trying to get a path to a Resource but I have had no luck.

This works (both in IDE and with the JAR) but this way I can\'t get a path to a file, only the file

Answer 1

You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:

src/main/resources/myfile.txt

If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.

Answer 2

follow code!

/src/main/resources/file

streamToFile(getClass().getClassLoader().getResourceAsStream("file"))

public static File streamToFile(InputStream in) {
    if (in == null) {
        return null;
    }

    try {
        File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
        f.deleteOnExit();

        FileOutputStream out = new FileOutputStream(f);
        byte[] buffer = new byte[1024];

        int bytesRead;
        while ((bytesRead = in.read(buffer)) != -1) {
            out.write(buffer, 0, bytesRead);
        }

        return f;
    } catch (IOException e) {
        LOGGER.error(e.getMessage(), e);
        return null;
    }
}

Answer 3

Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow

@Autowired
private ResourceLoader resourceLoader;

inside tour function write the bellow code in order to load file:

    Resource resource = resourceLoader.getResource("classpath:imports.xml");
    try{
        File file;
        file = resource.getFile();//will load the file
...
    }catch(IOException e){e.printStackTrace();}

Answer 4

Maybe this method can be used for quick solution.

public class TestUtility
{ 
    public static File getInternalResource(String relativePath)
    {
        File resourceFile = null;
        URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
        String codeLocation = location.toString();
        try{
            if (codeLocation.endsWith(".jar"){
                //Call from jar
                Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
                resourceFile = path.toFile();
            }else{
                //Call from IDE
                resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
            }
        }catch(URISyntaxException ex){
            ex.printStackTrace();
        }
        return resourceFile;
    }
}

Answer 5

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!");
}

Answer 6

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()

References:

getResource(), toExternalForm()