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.
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:
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：
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.
Answer：
In Spring Boot we have provision to override properties as below
mvn spring-boot:run -Dspring-boot.run.arguments=--server.port=8082