Spring: overriding one application.property from command line

To update a little things, the Spring boot 1.X Maven plugin relies on the --Drun.arguments Maven user property but the Spring Boot 2.X Maven plugin relies on the -Dspring-boot.run.arguments Maven user property.

So for Spring 2, you need to do :

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081"

And if you need to pass multiple arguments, you have to use , as separator and never use whitespace between arguments :

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081,--foo=bar"

About the the maven plugin configuration and the way of passing the argument from a fat jar, it didn't change.
So the very good Andy Wilkinson answer is still right.

You can pass in individual properties as command-line arguments. For example, if you wanted to set server.port, you could do the following when launching an executable jar:

java -jar your-app.jar --server.port=8081

Alternatively, if you're using mvn spring-boot:run with Spring boot 2.x:

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081"

Or, if you're using Spring Boot 1.x:

mvn spring-boot:run -Drun.arguments="--server.port=8081"

You can also configure the arguments for spring-boot:run in your application's pom.xml so they don't have to be specified on the command line every time:

<plugin>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-maven-plugin</artifactId>
    <configuration>
        <arguments>
            <argument>--server.port=8085</argument>
        </arguments>
    </configuration>
</plugin>