how to pass application.properties in commandLine for a spring boot application?

Question

I have a spring boot application and I want to pass application.properties file in commandLine when I start-up.

i.e when I run mvn spring-boot:run --application.properties

I will have a default application.properties in src/main/resources. but that is only for testing purposes. In the production run, I would like to pass the property file in commandLine.

I am aware of passing single arguments such as

mvn spring-boot:run --server.port=9001.

But I have many such properties and would prefer to pass a property file if that is possible.

Answer 1

You can do that with spring.config.location property:

mvn spring-boot:run -Dspring.config.location=your.properties

Answer 2

In case if anyone finds it useful as it was for me. If you want to pass in individual application properties as parameters when using the maven spring boot run command you can use the argument spring-boot.run.jvmArguments.

Eg:

mvn spring-boot:run -Dspring-boot.run.jvmArguments='
-Dspring.datasource.url=jdbc:postgresql://localhost:5432/mydb 
-Dspring.datasource.username=admin 
-Dspring.datasource.password=admin'

With the above command I am setting (overriding) the following properties which were in the application.properties file.

spring.datasource.url=jdbc:postgresql://localhost:5432/mydb
spring.datasource.username=admin
spring.datasource.password=admin

Answer 3

mvn spring-boot:run -Dspring-boot.run.arguments=--spring.config.location=classpath:/application-local.properties