how to package the springboot application with dynamic end points on application.properties file - spring

Question

I have 1 spring boot application. In this application i have configured 1 eCommerce system as elastic path(Configure the end point url of elastic path in application.properties file). Now i have to give my spring boot application to some other guys. which will be deploy on tomcat server. I don't want to give the source code. So i can make the war file but now problem is that they have their own elastic path eCommerce and they wants to configure their own eCommerce.
I want to externalize some properties which will override the existing property.
My springboot application have 2 modules :
1) elasticpath module which is having elasticpath-application.properties
2) salesforce - salesforce-application.properties
Now i have to externalize "C:\apache-tomcat-8.5.29\conf\ep-external.properties" file which will override the existing property. Now problem is that #PropertySource is loading in last position. So my external file is not able to override the property.
#SpringBootApplication
#PropertySource(value = {"classpath:application.properties", "classpath:elasticpath-application.properties", "classpath:salesforce-application.properties")
public class SpringBootDemo extends SpringBootServletInitializer implements CommandLineRunner {
private static final Logger LOG = LoggerFactory.getLogger(SpringBootDemo.class);
private ServletContext servletContext;
#Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
//application = application.properties("file:C:\\apache-tomcat-8.5.29\\conf\\ep-external.properties");
return application.sources(SpringBootDemo.class);
}
#Override
public void onStartup(ServletContext servletContext) throws ServletException {
this.servletContext = servletContext;
super.onStartup(servletContext);
}
public static void main(String[] args) {
SpringApplication.run(SpringBootDemo.class, args);
}
#Override
public void run(String... args) throws Exception {
}
}

Answer 1

Yes absolutely possible. Basically what you need is changing the property value as needed without changing jar/war
Passing command line args for jar
Package your spring boot application as jar and put the external application.properties file at any location and pass on the same location as command line argument as below :
java -jar app.jar --spring.config.location=file:<property_file_location>
This will pick up the external properties.
Passing command-line/dynamic args for war
1. Extend the SpringBootServletInitializer as below
#SpringBootApplication
class DemoApp extends SpringBootServletInitializer {
private ServletContext servletContext;
public static void main(String[] args){SpringApplication.run(DemoApp.class,args);}
#Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
builder = builder.properties("test.property:${test_property:defaultValue}");
return builder.sources(DemoApp.class)
}
#Override
public void onStartup(ServletContext servletContext) throws ServletException {
this.servletContext = servletContext;
super.onStartup(servletContext);
}
}
Access the property as usual like below anywhere you want :
#Value("${test.property}")
Before starting tomcat set the env variable named test_property. that's it
Additionally:
You can pass on the property like below as well if you want to supply complete file as external file.
.properties("spring.config.location:${config:null}")
Further reading about externalized configuration : https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html