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Spring Profile Based Jersey Rest Service publish

It is possible to publish jersey rest service based on spring profile? lets say as following example, how can I publish RegisterServices1 when using profile1?

public class ApiGWRestApplicationConfig extends ResourceConfig {

   public ApiGWRestApplicationConfig() {     
      register(RegisterServicesApiGWInterface.class);
    }
}

@Service
@Profile("profile1")
@Path(SystemConstants.REST_REGISTER)
public class RegisterServices1 implements RegisterServicesApiGWInterface {

}

@Service
@Profile("profile2")
@Path(SystemConstants.REST_REGISTER)
public class RegisterServices2 implements RegisterServicesApiGWInterface{}

web.xml

<servlet>
    <servlet-name>jersey-servlet-kagw</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>javax.ws.rs.Application</param-name>
        <param-value>com.ttech.tims.imos.web.ApiGWRestApplicationConfig</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
about 4 years ago · Santiago Trujillo
1 Respuestas
Responde la pregunta

0

So what you can do is get a hold of the ApplicationContext and use getBeansWithAnnotation(Path.class). This will give you all the resource instances that are part of the profile. Then you can register the instances.

I though it would be possible to inject the ApplicationContext into the ResourceConfig, but as mentioned in the comment above, it seems the creation of the ResourceConfig doesn't have access to it yet.

What I was able to get to work, is to use a JAX-RS Feature which also has access to registration methods, just like you have in the ResourceConfig. Using the Feature will give you access to the ApplicationContext

public class SpringProfilesFeature implements Feature {

    @Inject
    private ApplicationContext context;

    @Override
    public boolean configure(FeatureContext featureContext) {
        Map<String, Object> resources = context.getBeansWithAnnotation(Path.class);

        resources.values().forEach(resource -> featureContext.register(resource));

        return true;
    }
}

Then just register the feature with the ResourceConfig

public AppConfig() {
    register(SpringProfilesFeature.class);
}

Remove any other registrations you have for all your resources. Just let the feature register them.

I've confirmed that this works. Not sure how you set your profile for the environment, but hopefully this is something you already know how to do.

about 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
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