You don't need to mess with class loaders. In fact it's a bad habit to get into because class loader resources are not java.io.File objects when they are in a jar archive.

Maven automatically sets the current working directory before running tests, so you can just use:

    File resourcesDirectory = new File("src/test/resources");

resourcesDirectory.getAbsolutePath() will return the correct value if that is what you really need.

I recommend creating a src/test/data directory if you want your tests to access data via the file system. This makes it clear what you're doing.

You need ServletContext.getRealPath(String) method.

getServletContext().getRealPath("/WEB-INF")

The problem was that I have not included the resource in the maven build in pom file. When I included the following it worked with abcManager.class.getResource("/abc.access")

<build>
    <resources>
      <resource>
                <directory>src/main/resources</directory>
                <includes>
                    <include>abc.access</include>
                </includes>
            </resource>
    </resources>   
</build>

Try this.

InputStream is = getClass().getClassLoader()
                         .getResourceAsStream("/words.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(is));

You'll be able to elicit the path of the resource in the JAR, but the running program can't know, where your development repository is, unless you supply the information explicitly.

Use class that is in parellel to the resource and use its getResourceAsStream method. For an example for following structure you can use below code.

package com.aptkode;

import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.StandardCharsets;

public class GetResourceInJar {
    public static void main(String[] args) {
    try(final InputStream stream = GetResourceInJar.class.getResourceAsStream("com/aptkode/resource.txt")) {
        if (stream != null) {
            final String text = new String(stream.readAllBytes(), StandardCharsets.UTF_8);
            System.out.println(text);
        } else {
            System.err.println("failed to read text");
        }
    }
    catch (Exception e)
    {
        System.err.println(e.getMessage());
    }
}
}

Inside jar, resource file will be as follows.

Running command java -jar .\target\get-resource-in-jar-1.0-SNAPSHOT.jar will print the content. Following is the maven jar plugin configuration.

<plugin>
    <groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-jar-plugin</artifactId>
    <version>3.2.0</version>
    <configuration>
        <archive>
            <manifest>
                <mainClass>
                    com.aptkode.GetResourceInJar
                </mainClass>
            </manifest>
        </archive>
    </configuration>
</plugin>

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.