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PropertiesLauncher command line arguments not working with Spring Boot executable Jar

So I have basic multi-module Spring Boot project. The goal, that I had was to build executable jar and pass additional properties with the help of -Dloader.path=....
For some reason (if I understand purpose of this argument) loader.path is being ignored completely.
My project structure is following:
\-
|--conf
|---default
|--pets-api
|--pets-app (this module contains the Main-Class)
|--pets-domain
|--pets-infrastructure
Since no custom active profile is being passed it uses "default". Jar contains application-default.propeties file, that has single configuration server.servlet.context-path=/v1.
/conf/default location has 2 properties files:
application.properties
random.properties - this is bind to #ConfigurationProperties(prefix = "...") inside application
When I run it normally everything is fine java -jar pets-app-0.0.1-SNAPSHOT.jar. It just uses application-default.properties file and that is it.
Now when I am trying to utilize -Dloader.path argument as in java -Dloader.path=PATH/TO/conf/default -jar pets-app-0.0.1-SNAPSHOT.jar it starts application same as before, as if I am not adding 2 more file to classpath.
What is used:
Java 17
Spring Boot 2.6.12
Gradle
Did anyone come across this as well? Any suggestion on how to resolve it?
PS. If there is need to see the code, I can upload it to GitHub.

Pass gradle command line arguments to application.properties

I have a Spring Boot Gradle application and many MySQL servers and databases.
In different scenarios, I want to start the application with different databases or create them if they do not exist. I want to use this mechanism with command line arguments.
The URL for database is stored in application.properties file of Spring:
spring.datasource.url=jdbc:mysql://localhost:3306/mydb1?createDatabaseIfNotExist=true&useSSL=true
I want to run gradle with command line arguments that will apply to the application.properties file so in my head will be something like this.
application.properties :
spring.datasource.url=jdbc:mysql://${linkNewDB}?createDatabaseIfNotExist=true&useSSL=true
And to run the program as :
gradle bootRun -Pargs=--linkNewDB="someNewDB:3309"
Does anybody know how can I achieve this mechanism? I tried different options but none of them worked. Thanks!

How about changing the whole URL, like this:
gradle bootRun --args='--spring.datasource.url=jdbc:mysql://someNewDB:3309?createDatabaseIfNotExist=true&useSSL=true'
reference: https://docs.spring.io/spring-boot/docs/current/gradle-plugin/reference/htmlsingle/#running-your-application.passing-arguments

Can I pass an internal Spring Boot variable to gradlew?

I have a question. In my app, I do have several configurations:
application-prod.properties
application-test.properties
application-dev.properties
and the main file:
application.properties
Containing one line:
spring.profiles.active=test
To build and run the app I am using ./gradlew buildNeeded
Can I somehow pass the properties suffix: test, prod, dev so it is used in the build process, so I can make different bash scripts to run the installation process on test and prod servers?
I am looking for something like ./gradlew buildNeeded --spring.profiles.active=test or anything that will work...

You can achieve that using spring-gradle-plugin.
From plugin documentation Passing arguments to your application
Like all JavaExec tasks, arguments can be passed into bootRun from the
command line using --args='' when using Gradle 4.9 or
later. For example, to run your application with a profile named dev
active the following command can be used:
./gradlew bootRun --args='--spring.profiles.active=dev'

You can use project properties:
task buildNeeded {
doLast {
def springProfile = project.properties.getOrDefault('spring.profiles.active', 'test')
println springProfile
}
}
Then you can run the project with
./gradlew buildNeeded -Pspring.profiles.active=test

How to edit app.properties file with build parameter in Jenkins

I have a selenium script which I want to run from Jenkins. I have a properties file called app.properties. This file consists properties such as:
browser=chrome
I configured my project as parameterized so when I run my build, it asks for browser parameter. I want to select this parameter(for example firefox), so that it will change browser property in app.properties and will run the automation in Firefox.
Normally, when I change the browser property in app.properties in Intellij, my program runs with that browser. So there is nothing wrong with my program in that sense.
Is there a way to change app.properties with respect to my Jenkins build parameter and run the program with that configuration?
EDIT: I found the following solution:
Install surefire plugin.
Add a browser parameter.
In your property managing class, take browser parameter as
System.getProperty("browser");
From jenkins, configure a browser parameter
Invoke maven command: mvn test "-Dbrowser=${BROWSER}"

You can pass system properties to change configuration.
First you should configure your project to read both system properties and configuration file, where system properties will have higher priority. I'd recommend Apache Commons Composite Configuration. It can look like this:
CompositeConfiguration configuration = new CompositeConfiguration();
try {
configuration.addConfiguration(new SystemConfiguration());
configuration.addConfiguration(new PropertiesConfiguration("app.properties"));
} catch (ConfigurationException e) {
e.printStackTrace();
}
//Read your configuration values here
This way when you provide system property -Dbrowser=chrome it will override value from configuration file.
Secondly, you'll need to configure Jenkins job. Since you're passing a parameter you can use it in your build step definition:
mvn clean test -Dbroswer=${browser-param}

“The way parameters take effect is also different depending on the parameter type you choose ... String parameters are exposed as environment variables of the same name.”
https://wiki.jenkins.io/plugins/servlet/mobile?contentId=34930782#content/view/34930782

Maven run configuration doesn't pick up active profile while Spring run configuration does

My Spring run configuration is just the default with the following the VM options:
-Dspring.profiles.active=local
My Maven run configuration is the defaults with the following in the command line:
spring-boot:run
and the following in VM options:
-Dspring.profiles.active=local
When I run the maven one it doesn't pick up on a profile and instead uses default:
No active profile set, falling back to default profiles: default
With the Spring configuration I have no issues, the problem is with how I need to deploy it, it uses a maven command, so I can't have this failing and I don't really understand why it's happening. There really isn't anything fancy is this projects. It's your basic micro service.
Just in case it's needed. The root 'Application' file only has the following:
#SpringBootApplication
public class MyApplication {
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
}
Link to the Maven run config. It is all defaults outside of what is pictured. https://imgur.com/a/GGiwimQ

Adding the following to the command line arguments seemed to fix it:
-Drun.jvmArguments=-Dspring.profiles.active=local
Although I understand why it worked it doesn't explain why this happened in the first place. I will update this answer if I ever find out the true reason.

create
application.properties
file in src/main/resources if it doesnt exist and add
spring.profiles.active=local