How to package config.yml and resources in $project--0.0.1-SNAPSHOT.jar - maven

Question

I've a drop wizard project which has the following directory structure:
basedir
pom.xml
config.yml
src
main
resources
myresource.xml
The build portion of
<build>
<finalName>project-${version}</finalName>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.1</version>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>2.1</version>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
<configuration>
<transformers>
<transformer implementation="org.apache.maven.plugins.shade.resource.ManifestResourceTransformer">
<mainClass>project-package.App</mainClass>
</transformer>
<transformer implementation="org.apache.maven.plugins.shade.resource.ServicesResourceTransformer">
</transformer>
</transformers>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
When I run my code from Eclipse, I use command as follows:
$mypackage.App server config.yml
It works fine and I can read the file under resources directory.
When I use mvn to build, in the snapshot jar created, it does not include config.yml and the path to "myresource.xml" is missing even though it is included in the snapshot.jar in the basedir. So, when I run it from my jar file, my code does not find it as resources.
How should I change the pom.xml to include config.yml and package the resource directory such a way that it is accessible as resource.

Answer 1

It is really strange that
this.getClass().getResource("myresource.xml")
does work in IDE but NOT when you run it from a jar.
this.getClass().getResourceAsStream("myresource.xml")
works both inside IDE and from jar.
That resolves the resources issue. However, I need an answer for how to package config.yml file. Should I supply separately in addition to the jar file?