Spring: overriding one application.property from command line

Question

I have an application.properties file with default variable values. I want to be able to change ONE of them upon running with mvn spring-boot:run. I found how to change the whole file, but I only want to change one or two of these properties.

Answer 1

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> 

Answer 2

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.