Decide which of multiple implementations to use in Spring Boot app

Question

Given two (or more) implementations of a particular service API, what's the best way to pick which one to use at runtime in my app based on an application property?

Example API:

public interface Greeting { String sayHello(String username); }

Implementations:

public class FriendlyGreeting implements Greeting { public String sayHello(String username) { return "Hello, " + username; } } public class HostileGreeting implements Greeting { public String sayHello(String username) { return "Go away, " + username; } }

I've got a separate service class with an @Autowired constructor that takes an instance of Greeting. What I want, is based upon a configuration property, to decide which greeting implementation gets injected and used. I came up with using a configuration class to make that decision:

@Configuration public class GreetingConfiguration { private String selection; @Autowired public GreetingConfiguration(@Value("${greeting.type}") String type) { this.selection = type; } @Bean public Greeting provideGreeting() { if ("friendly".equals(selection)) { return new FriendlyGreeting(); } else { return new HostileGreeting(); } } }

Is this the right way to do what I want? I went down the road of using @Qualifier on the implementations, and ended up with a mess where Spring saw 3 instances of my Greeting API, and I needed a configuration anyway to pick which implementation to use and return it with a unique qualifier name on it, and that feels worse than what I settled on.

Does it have to be a configuration property or can it be a Spring profile? Because using a Profile would be the nice and calm Springy way to do this. After that, depending on profile, you would provide either Hostile or Friendly Configuration to handle the bean. — Compass
– Compass, Commented Jan 23, 2019 at 20:18
Specifically: have suitable @Profile annotations on your components and pass the spring.profiles.active and/or spring.profiles.include property at runtime... — user268396
– user268396, Commented Jan 23, 2019 at 20:22
I'll look into profiles, but I think they're defined by environment that we deploy the app into and not really changeable. — David
– David, Commented Jan 23, 2019 at 20:22
Also possibly helpful is using @Conditional stackoverflow.com/a/34351004/2958086 — Compass
– Compass, Commented Jan 23, 2019 at 20:29

Vitor Gabriel Carrilho · Accepted Answer · 2019-01-23 21:39:13Z

2

You can mark both Greeting as @Service and select the chosen one with @Qualifier("yourServiceHere") like this:

@Autowired @Qualifier("friendlyGreeting") private Greeting greeting;

Another way you can do it is with profile. You can mark your FriendlyGreeting service with @Service and @Profile("friendly") and the HostileGreeting service with @Service and @Profile("hostileGreeting") and just put in the application.properties the following:

spring.profiles.active=friendly

edited Jan 23, 2019 at 21:39

answered Jan 23, 2019 at 20:57

Vitor Gabriel Carrilho

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4 Comments

xerx593 Over a year ago

i doubt, that SpEL works in qualifier... (stackoverflow.com/a/29773117/592355)

David Over a year ago

Good thought, but the evaluation of "${app.greeting}" doesn't happen, and spring goes looking for a bean with the literal qualifier name of ${app.greeting}

Vitor Gabriel Carrilho Over a year ago

So you can put your qualifier as a string @Qualifier("friendlyGreeting")

Fuad Efendi Over a year ago

so you cannot manage which implementation to use at runtime without using Profile: Qualifier("friendlyGreeting") is hardcoded; and Qualifier("${app.greetig}") won't work. Is that correct, Profile is the only way (except initial solution in the question)?

David · Accepted Answer · 2019-01-30 17:33:35Z

Answering my own question.

@Compass and @user268396 were correct - using Profiles got this working as expected.

I created both implementations, annotated with @Service and @Profile("friendly") or @Profile("hostile"), and could change the property spring.profiles.active to dev,friendly for example, and get what I wanted.

Fuad Efendi · Accepted Answer · 2020-12-07 21:02:42Z

You can use @Conditional annotations described at https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/context/annotation/Conditional.html and https://reflectoring.io/spring-boot-conditionals/

@Profile annotations mentioned above are based upon @Conditional(from Spring Framework); see also Spring Boot: org.springframework.boot.autoconfigure.condition

toddcscar · Accepted Answer · 2021-05-25 20:36:52Z

Here is a full solution using ideas mentioned by David and Vitor above with @Profile and @Qualifer annotations.

Two beans with same name but Only one is activated based on which profile is defined.

@Profile("profile1") @Bean("greeting") public class FriendlyGreeting implements Greeting { --- @Profile("profile2") @Bean("greeting") public class HostileGreeting implements Greeting { --- @Configuration public class GreetingConfiguration { private Greeting greeting; @Autowired public GreetingConfiguration(@Qualifier("greeting") Greeting greeting) { this.greeting = greeting; } }

Notes:

you can remove the intermediate class GreetingConfiguration and stick the "greeting" bean wherever you need
i prefer the @Autowired on the constructor instead of the class member to make it easier for unit testing.

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