Executable jar won't find the properties files

BalusC is right, you need to instruct Maven to generate a MANIFEST.MF with the current directory (.) in the Class-Path: entry.

Assuming you're still using the Maven Assembly Plugin and the jar-with-dependencies descriptor to build your executable JAR, you can tell the plugin to do so using the following:

  <plugin>
    <artifactId>maven-assembly-plugin</artifactId>
    <version>2.2</version>
    <configuration>
      <descriptorRefs>
        <descriptorRef>jar-with-dependencies</descriptorRef>
      </descriptorRefs>
      <archive>
        <manifest>
          <mainClass>com.stackoverflow.App</mainClass>
        </manifest>
        <manifestEntries>
          <Class-Path>.</Class-Path> <!-- HERE IS THE IMPORTANT BIT -->
        </manifestEntries>
      </archive>
    </configuration>
    <executions>
      <execution>
        <id>make-assembly</id> <!-- this is used for inheritance merges -->
        <phase>package</phase> <!-- append to the packaging phase. -->
        <goals>
          <goal>single</goal> <!-- goals == mojos -->
        </goals>
      </execution>
    </executions>
  </plugin>

There are two workarounds:

1. Don't use the JAR as executabele JAR, but as library.

   java -cp .;filename.jar com.example.YourClassWithMain
   

2. Obtain the root location of the JAR file and get the properties file from it.

   URL root = getClass().getProtectionDomain().getCodeSource().getLocation();
   URL propertiesFile = new URL(root, "filename.properties");
   Properties properties = new Properties();
   properties.load(propertiesFile.openStream());
   

None of both are recommended approaches! The recommend approach is to have the following entry in JAR's /META-INF/MANIFEST.MF file:

Class-Path: .

Then it'll be available as classpath resource the usual way. You'll really have to instruct Maven somehow to generate the MANIFEST.MF file like that.