Access src/main/resources/ dynamically from source code

Question

My files are located under projectRoot/src/main/resources/levels/

enter image description here

If I call Utils.getFileNamesInDirectory("src/main/resources/levels"), it works.
But when project is packaged, levels directory is placed under root of .jar.
How can I made this dynamic in static class? i.e src/main/resources/
So that code will run within eclipse and as standalone jar.

Code to list files in directory ..

public class Utils {
    public static List<String> getFileNamesInDirectory(String directory){

        List<String> results = new ArrayList<String>();

        File[] files = new File(directory).listFiles(new FilenameFilter() {
            public boolean accept(File dir, String filename) {
                return filename.endsWith(".json");
            }
        });

        for (File file : files) {
            if (file.isFile()) {
                results.add(file.getName());
            }
        }

        Collections.sort(results);

        return results;
    }
}

Updated

I've moved to using getResourceAsStream (as getResource was causing IllegalArgumentException: URI is not hierarchical) and I'm able to list files in a directory within Eclipse.

public static List<String> getFileNamesInDirectory(String directory){

        List<String> results = new ArrayList<String>();
        InputStream in = Utils.class.getResourceAsStream("/" + directory);
        BufferedReader rdr = new BufferedReader(new InputStreamReader(in));
        String line;

        try {

            while ((line = rdr.readLine()) != null) {
                System.out.println("file: " + line);
                results.add(new File(line).getName());
            }

            rdr.close();
        } catch (IOException e1) {
            e1.printStackTrace();
        }

        Collections.sort(results);

        return results;
    }

But when I run it as standalone .jar I get the following error on this line: while ((line = rdr.readLine()) != null) {
Why does it not work outside Eclipse?

Exception in thread "main" java.lang.NullPointerException
    at java.io.FilterInputStream.read(FilterInputStream.java:133)
    at sun.nio.cs.StreamDecoder.readBytes(StreamDecoder.java:322)
    at sun.nio.cs.StreamDecoder.implRead(StreamDecoder.java:364)
    at sun.nio.cs.StreamDecoder.read(StreamDecoder.java:210)
    at java.io.InputStreamReader.read(InputStreamReader.java:205)
    at java.io.BufferedReader.fill(BufferedReader.java:165)
    at java.io.BufferedReader.readLine(BufferedReader.java:328)
    at java.io.BufferedReader.readLine(BufferedReader.java:393)
    at com.app.tools.Utils.getFileNamesInDirectory(Utils.java:31)

You don't. `/src/main/resources` is a **project** directory convention used by Maven to contain resources that will end up at the root of a generated archive/classpath. — Sotirios Delimanolis, Aug 31 '14 at 18:55
In, eclipse you should also use the classpath rather than the file system. — Sotirios Delimanolis, Aug 31 '14 at 18:56
I'm aware `src/main/resources` is a directory convention for `Maven`. Do you have example of classpath use over filesystem? — bobbyrne01, Aug 31 '14 at 19:02
See http://docs.oracle.com/javase/7/docs/api/java/lang/Class.html#getResourceAsStream%28java.lang.String%29. — Kevin Krumwiede, Aug 31 '14 at 19:03

score 3 · Accepted Answer · edited May 23 '17 at 10:26

3

The reason your code works in Eclipse is that Eclipse launches the java process from the project directory and we can assume the path provided in

Utils.getFileNamesInDirectory("src/main/resources/levels")

is relative to the current working directory (the project directory). Since the file system location <project-directory>/src/main/resources/levels exists, it can be found and returned to you.

src/main/resources is a Maven convention meant to hold resources that will eventually end up in the classpath when the project is compiled/built/deployed. To retrieve resources from the classpath you use Class#getResource(String), ClassLoader#getResource(String) and/or ClassLoader#getSystemResource(String).

Now, although there are ways to list resources in the classpath, you should not typically do this. If you need a resource, you know it by name and can therefore use one of the methods listed above to get it.

edited May 23 '17 at 10:26

Community

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answered Aug 31 '14 at 19:11

Sotirios Delimanolis

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I'm no longer using file system approach, instead using `getResourceAsStream` but still having issues, any ideas? updated question – bobbyrne01 Aug 31 '14 at 23:00
@bobbyrne01 `getResourceAsStream` returns `null` if no matching resource can be found. Does your `.jar` contain the resource at the right (specified) path? – Sotirios Delimanolis Aug 31 '14 at 23:16
The .jar has folder at `/levels` i.e the root of .jar. I'm passing `/levels` to `getResourceAsStream` – bobbyrne01 Aug 31 '14 at 23:24
@bobbyrne01 A directory does not constitute a resource. Is `/levels` a directory or a file entry in the `.jar`? – Sotirios Delimanolis Aug 31 '14 at 23:32
`/levels` is a directory contains 3 files i.e `level1.json`, `level2.json` and `level3.json` .. all of which are in the .jar – bobbyrne01 Sep 01 '14 at 00:20
@bobbyrne01 So, you can access `/levels/level1.json` and the others as resources. – Sotirios Delimanolis Sep 01 '14 at 00:28
But I first need to list the filenames before I can access them, which is failing stated above, as the filenames are dynamic. – bobbyrne01 Sep 01 '14 at 09:52
@bobbyrne01 See the link in my answer. But you really shouldn't be accessing resources dynamically from the classpath. – Sotirios Delimanolis Sep 01 '14 at 18:03

Access src/main/resources/ dynamically from source code

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