在我看来，这是一个很好的解决方案:

interface PersonRepository extends Repository<Person, UUID> {

    <T> Collection<T> findByLastname(String lastname, Class<T> type);
}

像这样使用它

void someMethod(PersonRepository people) {

  Collection<Person> aggregates =
    people.findByLastname("Matthews", Person.class);

  Collection<NamesOnly> aggregates =
    people.findByLastname("Matthews", NamesOnly.class);
}

使用Spring Data JPA有一个从数据库中选择特定列的规定

----在DAOImpl ----

@Override
    @Transactional
    public List<Employee> getAllEmployee() throws Exception {
    LOGGER.info("Inside getAllEmployee");
    List<Employee> empList = empRepo.getNameAndCityOnly();
    return empList;
    }

----在回购----

public interface EmployeeRepository extends CrudRepository<Employee,Integer> {
    @Query("select e.name, e.city from Employee e" )
    List<Employee> getNameAndCityOnly();
}

这对我来说是100%有效的。 谢谢。

使用本机查询:

Query query = entityManager.createNativeQuery("SELECT projectId, projectName FROM projects");
List result = query.getResultList();

Use:

@Query("SELECT e FROM #{#entityName} e where e.userId=:uid")
List<ClienteEnderecoEntity> findInfoByUid(@Param("uid") UUID uid);

在我的情况下，我只需要json结果，这对我来说是有效的:

public interface SchoolRepository extends JpaRepository<School,Integer> {
    @Query("select s.id, s.name from School s")
    List<Object> getSchoolIdAndName();
}

控制器:

@Autowired
private SchoolRepository schoolRepository;

@ResponseBody
@RequestMapping("getschoolidandname.do")
public List<Object> getSchool() {
    List<Object> schools = schoolRepository.getSchoolIdAndName();
    return schools;
}

{
   "Comments":"Why not using JDBCTemplate",
   "Url":"https://www.baeldung.com/spring-jdbc-jdbctemplate"
}