Adding Spring (Boot?) to existing RESTEasy JAX-RS application - spring

Question

I have an existing Maven project based on JAX-RS using RESTEasy. It works great! It creates a WAR that I deploy to Tomcat running on Ubuntu. It's clean and follows the latest standards. The POM is simple:
...
<packaging>war</packaging>
...
<dependencies>
...
<dependency>
<groupId>javax.ws.rs</groupId>
<artifactId>javax.ws.rs-api</artifactId>
<version>2.0.1</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>org.jboss.resteasy</groupId>
<artifactId>resteasy-jaxrs</artifactId>
<version>3.1.0.Final</version>
</dependency>
<dependency>
<groupId>org.jboss.resteasy</groupId>
<artifactId>resteasy-servlet-initializer</artifactId>
<version>3.1.0.Final</version>
</dependency>
<dependency>
<groupId>org.jboss.resteasy</groupId>
<artifactId>resteasy-jackson2-provider</artifactId>
<version>3.1.0.Final</version>
</dependency>
...
<build>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.5.1</version>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-war-plugin</artifactId>
<version>3.0.0</version>
</plugin>
</plugins>
...
I don't need any web.xml because I'm using the latest Java EE annotations:
import javax.ws.rs.ApplicationPath;
import javax.ws.rs.core.Application;
#ApplicationPath("/")
public class MyRESTApplication extends Application {
final FooResource fooResource = new FooResourceService();
...
#Override
public Set<Object> getSingletons() {
return ImmutableSet.of(fooResource);
}
}
This is all simple and it's working so great! Now I just want a way to easily change FooResource implementations based on the profile --- in other words, wire my singletons. That's where Spring comes in, right? And I've been told that Spring Boot makes Spring even more awesome, and you can use it with anything, and it gives you an actuator that allows you to gain real-time inside on the health of your system.
Unfortunately all the Spring Boot books and tutorials seem to think I'm starting with one of their quick-start applications. But I already have a great, simple application. I just want to:
Get my application wiring, based on profiles, from an external configuration file (not annotations) via Spring.
Get whatever other goodness comes from Spring Boot, because apparently it is awesome and will completely transform my application.
How do I add Spring (or Spring Boot) to this simple little JAX-RS application?

Answer 1

we solved it that way, that we created a singleton spring bean, let's call it ServiceStartupClass, where we register all JAX-RS services.
Here some code snippet how we start our services:
import javax.ws.rs.ApplicationPath;
import org.glassfish.jersey.server.ResourceConfig;
#Component
#ApplicationPath("/")
public class ServiceStartupClass extends ResourceConfig {
#PostConstruct
public void startup() {
register(FooResource.class);
...
}
}
If you need any further help, let me know