Spring Test: Create Few Classes Inside Test - spring

Question

I have an interface without implementation classes.
public interface MyInterface {
String getName();
}
And I have another class which have MyInterface as dependency.
#Component
public class SomeClass {
#Autowired
private MyInterface implementation;
public MyInterface someMethod(List<MyInterface> list){
//logic
}
}
I have to test SomeClass, but I don't have MyInterface implementation classes. Is it possible to create in test few MyInterface implementation classes and add them to ApplicationContext when I can get them as spring beans with #Autowired?

Answer 1

Assuming you use JUnit 5, this is possible using #TestConfiguration and provide an implementation there:
import org.junit.jupiter.api.Test;
import org.junit.jupiter.api.extension.ExtendWith;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.test.context.TestConfiguration;
import org.springframework.context.annotation.Bean;
import org.springframework.test.context.junit.jupiter.SpringExtension;
#ExtendWith(SpringExtension.class)
class SomeServiceTest {
#Autowired
private MyInterface myInterface;
#TestConfiguration
static class TestConfig {
#Bean
public MyInterface myInterface() {
/**
* Long version of implementing your interface:
*
* return new MyInterface() {
* #Override
* public String getName() {
* return null;
* }
* };
*/
// shorter ;-)
return () -> "Hello World!";
}
}
#Test
void test() {
SomeService someService = new SomeService(myInterface);
System.out.println(someService.doFoo());
}
}
But you could also use Mockito and create a mock for your MyInterface and have a fast unit test:
#ExtendWith(MockitoExtension.class)
class SomeServiceTest {
#Mock
private MyInterface myInterface;
#InjectMocks
private SomeService someService;
#Test
void test() {
when(myInterface.getName()).thenReturn("Test");
System.out.println(someService.doFoo());
}
}
While the code above technically answers what you asked for, always consider, as J Asgarov mentioned in the comments, if it makes sense to do it in this way.