Java: Relative path to external text file

Question

I am a bit confused with path in Java (using Eclipse). This is my file structure:

 Folder

    Subfolder

        file.txt

    jarfile.jar

So, I am trying to make the jar file parse data from file.txt and I use the following code:

Scanner in = new Scanner(this.getClass().getResourceAsStream("./Subfolder/file.txt"));

I have made a runnable jar file with Eclipse, put it in the Folder, but it does not work. What is it that I am doing wrong?

Answer 1

Since you use a resource file via a Class object, the path to the resource must be absolute:

getClass().getResourceAsStream("/Subfolder/file.txt");

Note that doing what you do is a bad idea, that is, opening a scanner on a resource which you don't have a reference to:

new Scanner(someInputStreamHere());

you have no reference to that input stream, therefore you cannot close it.

What is more, .getResource*() return null if the resource does not exist; in this case you'll get an NPE!

Recommended if you use Java 6 (using Guava's Closer):

final URL url = getClass().getResource("/path/to/resource");

if (url == null) // Oops... Resource does not exist
    barf();

final Closer closer = Closer.create();
final InputStream in;
final Scanner scanner;

try {
    in = closer.register(url.openStream());
    scanner = closer.register(new Scanner(in));
    // do stuff
} catch (IOException e) {
    throw closer.rethrow(e);
} finally {
    closer.close();
}

If you use Java 7, just use a try-with-resources statement:

final URL url = getClass().getResource("/path/to/resource");

if (url == null) // Oops... Resource does not exist
    barf();

final InputStream in;
final Scanner scanner;

try (
    in = url.openStream();
    scanner = new Scanner(in);
) {
    // do stuff
} catch (IOException e) {
    // deal with the exception if needed; or just declare it at the method level
}