Question

how can I use resources from other maven modules? My goal is to provide a AbstractImportClass as well as the to be imported files in a specific maven module. And use this module within other modules extending this class.

Let’s say ModuleA contains src/main/java/MyAbstractImportClass.java, and src/main/resources/MyImport.csv

I now want to use the abstract import class in ModuleB. Or rather, I will extend it, use the abstract-fileimport, and a few custom functions.
Then ModuleC also uses the abstracts’ import and some custom functions.

The problem is: the import in abstract class goes with reader and InputStream. When I execute just ModuleA everything is fine.

But when I tried to include the module via maven pom, and then extend the module to call the import, then I get NullPointerException at the line where the reader is used.

So obvious I cannot use foreign module resources this way.

But how could I instead make use of this?

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Update:

Module A:

src/main/java/path/to/MyClassA.java
src/main/resources/path/to/test.txt

abstract class MyClassA {
    public static String TESTFILE = test.txt;

    List<String> doImport(String filename) {
        InputStream fileStream = resourceClass.getResourceAsStream(filename);
        //some precessing
        return list;
    }
}

Module B:

src/main/java/path/to/MyClassB.java

class MyClassB implements MyClassA {
    List<String> list = doImport(TESTFILE);
}

If I put MyClassB in same dir as A, then everything works fine.
If I build B in a own module I get NullPointer for InputStream, what means the file is not found.

Answer 1

I don’t think your problem is related to Maven at all. Class.getResourceAsStream() resolves relative paths as relative to the class object that you call it on. Therefore, if you use that method in an abstract class, every subclass of it could be looking for the resource in a different place.

For example, given three classes:

Super:

package com.foo;
public class Super {
    { System.out.println(getClass().getResourceAsStream("test.properties")); }
}

Sub1, a subclass of Super:

package com.foo.bar;
import com.foo.Super;
public class Sub1 extends Super {}

Sub2, another subclass:

package com.foo.bar.baz;
import com.foo.Super;
public class Sub2 extends Super {}

If you create a Super, it’ll look for the classpath resource “/com/foo/test.properties” because that’s how the path “test.properties” resolves relative to the class com.foo.Super. If you create a Sub1, it’ll look instead in “/com/foo/bar/test.properties”, and for a Sub2 instance, it’ll look in “/com/foo/bar/baz/test.properties”.

You might want to use an absolute path to the resource instead of a relative one, or else have the subclasses specify paths relative to themselves. It depends on your design and what kind of abstraction you’re trying to achieve.