Spring Boot: Is it possible to use external application.properties files in arbitrary directories with a fat jar?

If you have not changed the defaults of Spring Boot (meaning you are using @EnableAutoConfiguration or @SpringBootApplication and have not changed any Property Source handling), then it will look for properties with the following order (highest overrides lowest):

1. A /config subdir of the current directory
2. The current directory
3. A classpath /config package
4. The classpath root

The list above is mentioned in this part of the documentation

What that means is that if a property is found for example application.properties under src/resources is will be overridden by a property with the same name found in application.properties in the /config directory that is "next" to the packaged jar.

This default order used by Spring Boot allows for very easy configuration externalization which in turn makes applications easy to configure in multiple environments (dev, staging, production, cloud etc)

To see the whole set of features provided by Spring Boot for property reading (hint: there is a lot more available than reading from application.properties) check out this part of the documentation.

As one can see from my short description above or from the full documentation, Spring Boot apps are very DevOps friendly!

It's all explained here in the docs:

http://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html

Which explains that this is the order of precedence:

• A /config subdir of the current directory.
• The current directory
• A classpath /config package
• The classpath root

It also points out that you can define additional properties files for overrides like so:

java -jar myproject.jar 
    --spring.config.location=classpath:/overrides.properties

If you use spring.config.location, then all the default locations for application.properties are also included. This means that you can set up default values in application.properties and override as required for a particular environment.

I managed to load an application.properties file in external path while using -jar option.

The key was PropertiesLauncher.

To use PropertiesLauncher, pom.xml file must be changed like this:

<build>
    <plugins>
        <plugin>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-maven-plugin</artifactId>
            <configuration>  <!-- added -->
                <layout>ZIP</layout> <!-- to use PropertiesLaunchar -->
            </configuration>
        </plugin>
    </plugins>
</build>

For this, I referenced the following StackOverflow question: spring boot properties launcher unable to use . BTW, In Spring Boot Maven Plugin document(http://docs.spring.io/spring-boot/docs/1.1.7.RELEASE/maven-plugin/repackage-mojo.html), there is no mention that specifying ZIP triggers that PropertiesLauncher is used. (Perhaps in another document?)

After the jar file had been built, I could see that the PropertiesLauncher is used by inspecting Main-Class property in META-INF/MENIFEST.MF in the jar.

Now, I can run the jar as follows(in Windows):

java -Dloader.path=file:///C:/My/External/Dir,MyApp-0.0.1-SNAPSHOT.jar -jar MyApp-0.0.1-SNAPSHOT.jar

Note that the application jar file is included in loader.path.

Now an application.properties file in C:\My\External\Dir\config is loaded.

As a bonus, any file (for example, static html file) in that directory can also be accessed by the jar since it's in the loader path.

As for the non-jar (expanded) version mentioned in UPDATE 2, maybe there was a classpath order problem.

You would be able to launch your spring boot appication with the external properties file path as follows:

java -jar {jar-file-name}.jar 
--spring.config.location=file:///C:/{file-path}/{file-name}.properties