I want to add an alternate entry point to my Spring-Boot application. I would prefer to keep this as a fat jar. Is this possible?
According to their documentation, the
I don't believe that property would apply in your case. There are 3 different "Launchers" (go back to the docs and see). If you are building a jar it uses the JarLauncher class. If you switch it to PropertiesLauncher then loader.main would be useful.
META-INF/MANIFEST.MF
Main-Class: org.springframework.boot.loader.PropertiesLauncher
I took a different approach and use a command line parameter to determine which class to use as my SpringApplication class. I only have a single main() method, but different Application classes with different configurations that are used based on a command line param.
I have a single class with a main() in it:
public static void main(String[] args) {
SpringApplication app;
if( ArrayUtils.contains(args, "--createdb")){
app = new SpringApplication(CreateDB.class);
args = (String[])ArrayUtils.add(args, "--spring.jpa.hibernate.ddl-auto=create");
} else {
app = new SpringApplication(Application.class);
}
app.setWebEnvironment(false);
app.setShowBanner(false);
app.addListeners(new ConfigurationLogger());
// launch the app
ConfigurableApplicationContext context = app.run(args);
// finished so close the context
context.close();
}
But I have 2 different SpringApplication classes: Application.class & CreateDB.class. Each class defines a different @ComponentScan
path as well as different @EnableAutoConfiguration
options and different @Configuration
options. Finally, based on my command line arguments, I can decide whether to programatically enable additional profiles/etc.
In my case, I want a different launcher to just create the DB schema and exit, so I've forced the command line parameter.
I would suggest having a single main
but using Spring profiles (or configuration properties) to select one or other "entry point" @Configuration
class.