如何注入JpaRepository的实现

Question

我想在UserService中使用UserRepository中的方法a，但我得到的是jpaRepository，而不是我的自定义实现，我应该如何编写类来获得它？

存储库：

@Repository
public interface UserRepository<UserEntity extends EntityInterface,Long> extends JpaRepository<UserEntity,Long> {
Optional<UserEntity> findUserByLogin(String login);
}

使用泛型方法的CrudAbstractService：

public abstract class CrudAbstractService<ENTITY extends EntityInterface, DTO extends DTOInterface> {
protected final JpaRepository<ENTITY, Long> jpaRepository;
protected final Validator<DTO> validator;
protected final MapperInterface<ENTITY, DTO> mapper;
private Class<ENTITY> entityClazz;
public CrudAbstractService(JpaRepository<ENTITY, Long> jpaRepository,
Validator<DTO> validator, MapperInterface<ENTITY, DTO> mapper) {
this.jpaRepository = jpaRepository;
this.validator = validator;
this.mapper = mapper;
}
public Iterable<DTO> findAll() {
List<ENTITY> allEntities = jpaRepository.findAll();
if (allEntities == null) {
throw new EntityNotFound(entityClazz);
}
List<DTO> mappedDTOs = mapper.toDTOs(allEntities);
return mappedDTOs;
}
public void delete(DTO dto) {
validator.validate(dto);
ENTITY entity = mapper.toEntity(dto);
jpaRepository.delete(entity);
}
public DTO save(DTO dto) {
validator.validate(dto);
ENTITY entity = mapper.toEntity(dto);
ENTITY save = jpaRepository.save(entity);
if (save == null) {
throw new EntityNotFound(entityClazz);
}
DTO mappedDTO = mapper.toDTO(save);
return mappedDTO;
}
}

CrudUserService的实现，我想在那里注入UserRepository而不是JpaRepository:

@Service
public class UserService extends CrudAbstractService<UserEntity,UserDTO> {
private MapperInterface<LectureEntity,LectureDTO> lectureMapper;
public UserService(UserRepository<UserEntity, Long> jpaRepository,
Validator<UserDTO> validator, MapperInterface<UserEntity, UserDTO> mapper,
MapperInterface<LectureEntity,LectureDTO> lectureMapper) {
super(jpaRepository, validator, mapper);
this.lectureMapper = lectureMapper;
}

public UserDTO findUserByLogin(String login) {
if (login == null) {
throw new UserNotFoundException();
}
//Here i want use UserRepository method instead of JpaRepository. 
Optional<UserEntity> userByLogin = jpaRepository.findUserByLogin(login);
UserEntity userEntity = userByLogin.orElseThrow(UserNotFoundException::new);
List<LectureEntity> reservations = userEntity.getReservations();
List<LectureDTO> lectureDTOS = lectureMapper.toDTOs(reservations);
UserDTO userDTO = mapper.toDTO(userEntity);
userDTO.setLectures(lectureDTOS);
return userDTO;
}
}

Answer 1

我认为您不需要使存储库接口通用
所以，替换这个：

@Repository
public interface UserRepository<UserEntity extends EntityInterface,Long> extends JpaRepository<UserEntity,Long> {
Optional<UserEntity> findUserByLogin(String login);
}

这个：

@Repository
public interface UserRepository extends JpaRepository<UserEntity,Long> {
Optional<UserEntity> findUserByLogin(String login);
}

并将其用于您的服务：

@Service
public class UserService extends CrudAbstractService<UserEntity,UserDTO> {
private MapperInterface<LectureEntity,LectureDTO> lectureMapper;
public UserService(UserRepository jpaRepository,
Validator<UserDTO> validator, MapperInterface<UserEntity, UserDTO> mapper,
MapperInterface<LectureEntity,LectureDTO> lectureMapper) {
super(jpaRepository, validator, mapper);
this.lectureMapper = lectureMapper;
}
}

如果你需要将你的实体映射到DTO，那么你可以尝试使用JPA投影

关于在findAll()中抛出异常，在我看来，这不是一个好主意。您可能应该只返回一个空列表，并让类的客户端决定在缺少实体的情况下该怎么做。

同样在您的情况下，我会尽量避免使用抽象类和继承，而是使用组合。继承与组合：如何选择？为什么我更喜欢组合而不是继承？