無法實體化型別為“UserServiceImpl”的名為“userService”的@InjectMocks欄位。您尚未提供實體

我遇到了一個名為“無法實體化名為 'userService' 型別的 'UserServiceImpl' 的 @InjectMocks 欄位的模擬測驗的問題。您沒有在欄位宣告時提供實體，所以我嘗試構建實體。” 我一直在通過 stackoverflow 進行搜索，但不幸的是我還沒有找到適合自己的解決方案。所以，我有介面'UserService'，它的實作和我想測驗的幾個方法。

用戶服務：

public interface UserService {
Optional<User> getByToken(String token);
PagingDto<User> getUsers(String filter, Integer startIndex, Integer count);}

用戶服務實作：

public class UserServiceImpl extends AbstractPagingService<User, Integer, UserEntity> implements UserService {
private final UserRepository userRepository;

@Autowired
public UserServiceImpl(ProviderConfiguration providerConfiguration, UserRepository userRepository) {
    super(providerConfiguration.getFilter().getMaxResult(), userRepository, Sort.by("surname"));
    this.userRepository = userRepository;
}
four methods

測驗類：

@ExtendWith(SpringExtension.class)
class UserServiceImplTest {
private static final String TOKEN = "token";

@InjectMocks
private UserServiceImpl userService;

@Spy
private UserRepository userRepository;

@Test
void testGetByToken() {
    final UserEntity userEntity = new UserEntity();
    final int id = 1;
    final User expectedUser = new User();
    final String email = "[email protected]";
    final String surname = "surname";
    final String name = "name";
    final String patronymic = "patronymic";

    userEntity.setId(id);

    expectedUser.setId(id);
    expectedUser.setEmail(email);
    expectedUser.setSurname(surname);
    expectedUser.setName(name);
    expectedUser.setPatronymic(patronymic);

    when(userRepository.findById(id)).thenReturn(Optional.of(userEntity));
    final User actualUser = userService.getByToken(TOKEN).orElseThrow();
    assertEquals(expectedUser, actualUser);
    assertEquals(email, userEntity.getEmail());
    assertEquals(surname, userEntity.getSurname());
    assertEquals(name, userEntity.getName());
    assertEquals(patronymic, userEntity.getPatronymic());
}

@Test
void testGetAbsentUserByToken() {
    assertFalse(userService.getByToken(TOKEN).isPresent());
}

@Test
void testEmptyUser() {
    when(userRepository.findById(any())).thenThrow(IllegalArgumentException.class);
    assertFalse(userService.getByToken(TOKEN).isPresent());
}

如果我沒有擴展 AbstractPagingService<User, Integer, UserEntity>，一切都很好。但如果我這樣做，總會有我上面提到的錯誤。

uj5u.com熱心網友回復：

您缺少一個模擬，ProviderConfiguration它是您的服務所需的依賴項。

@ExtendWith(SpringExtension.class)
class UserServiceImplTest {
  private static final String TOKEN = "token";

  @InjectMocks
  private UserServiceImpl userService;

  @Spy
  private UserRepository userRepository;

  @Mock
  private ProviderConfiguration providerConfiguration;

可能需要更多步驟才能正確設定，包括getFilter()模擬或通過模擬回傳模擬（通常不鼓勵）。

如果ProviderConfiguration是一個沒有太多邏輯的資料類，請考慮在您的測驗中分配一個真實的配置：

@ExtendWith(SpringExtension.class)
class UserServiceImplTest {
  private static final String TOKEN = "token";

  private ProviderConfiguration providerConfiguration = createTestProviderConfiguration();

  @InjectMocks
  private UserServiceImpl userService;

  @Spy
  private UserRepository userRepository;

  private static ProviderConfiguration createTestProviderConfiguration() {
    final Filter filter = new Filter();
    filter.setMaxResult(42);

    final ProviderConfiguration cfg = new ProviderConfiguration();
    cfg.setFilter = filter;
    return cfg;
  }

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