Junit for Spring Cache Manager

developer picture developer · Sep 13, 2016 · Viewed 10.3k times · Source

I am using spring caching layer in my application and I have got an issue as part of writing unit tests to test spring cache layer using Mockito.

Please refer the below code for my problem:

Service layer:

    public CustomerServiceImpl implements CustomerService {
    @Autowired
    private CacheManager cacheManager;

    @Autowired
    private CustomerRepository customerRepository;//Repository is simple JPA repository interface which contains findByCustomerName()

    @Override
    @CachePut(value = "#customer", key = "#customer.customerName")
    public Customer insertOrUpdate(Customer customer) {
        return customerRepository.save(customer);
    }

    @Cacheable(value="customersCache", key = "#customerName")
    public Customer findByCustomerName(String customerName) {
        Customer customer = customerRepository.findByCustomerName(customerName);
        return customer;
    }
}

JUnit test Code for Service layer:

@RunWith(PowerMockRunner.class)
@PrepareForTest(CustomerServiceImplTest.class)
public class CustomerServiceImplTest {

    @Spy
    CacheManager cacheManager = new ConcurrentMapCacheManager("customersCache");

    @Mock
    CustomerRepository CustomerRepository;

    @InjectMocks
    CustomerServiceImpl customerServiceImpl = new CustomerServiceImpl();

    @Before
    public void setup() {
        MockitoAnnotations.initMocks(this);
    }

    @Test
    public void testCacheForFindByCustomerName() {
        Customer customer1 = new Customer();
        customer1.setId("1");
        customer1.setName("John");
        Customer customer2 = new Customer();
        customer2.setId("2");
        customer2.setName("Sara");

        //this should save to cache   
        Mockito.when(CustomerRepository.save(customer1))
        .thenReturn(customer1);
        customerServiceImpl.insertOrUpdate(customer1);

        //Now it should retreive from cache, but not able to
      Mockito.when(CustomerRepository.findByCustomerName(Mockito.any(String.class)))
                .thenReturn(customer1, customer2);

        Customer result = customerServiceImpl.findByCustomerName("John");
        assertThat(result, is(customer1));

        result = customerServiceImpl.findByCustomerName("John");
        assertThat(result, is(customer1));
    }
}

Exception:

I am getting an "java.lang.AssertionError:" because the caching layer did not work and the call has been passed to repository object (twice) which has returned the 'customer2' mock object above i.e., the repository method has been invoked twice for the same key by passing the service layer.

Also, please note that I am using "Mockito" framework for my tests.

I have tried to google for unit tests on spring caching and also referred the below URL, which is almost using the same concept, but it does not work for my above code.

How to test Spring's declarative caching support on Spring Data repositories?

Could you please help to resolve the above exception ?

Answer

Sean Patrick Floyd picture Sean Patrick Floyd · Sep 14, 2016

The Spring Cache Manager relies on Spring managing the application. You can't get that with the PowerMockRunner, you need to use SpringJUnit4Runner. You can still use PowerMock or Mockito programmatically, but not as a Runner.

Typically, you'll turn your unit test into a Spring-style integration test, something like this:

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration
public class SpringTest {

    @Configuration
    @EnableCaching
    static class SpringConfig{
        @Bean
        public CustomerService customerService(){
            return new CustomerServiceImpl(customerRepository());
        }
        @Bean
        public CustomerRepository customerRepository(){
            return Mockito.mock(CustomerRepository.class);
        }
    }

    @Autowired
    CustomerService customerService; // this will contain a proper managed cache

    @Autowired
    CustomerRepository customerRepository; // this is a mockito mock you can fine-tune
}