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;
}

你可以更新你的JPARepository，如下所示。

@Query("select u.status from UserLogin u where u.userId = ?1 or u.email = ?1 or u.mobile = ?1")
public UserStatus findByUserIdOrEmailOrMobile(String loginId);

UserStatus为Enum

public enum UserStatus
{
    New,
    Active,
    Deactived,
    Suspended,
    Locked
}

你可以使用@jombie给出的答案，并且:

将接口放在实体类之外的单独文件中; 是否使用本机查询(选择取决于您的需要); 不要为此重写findAll()方法，而是使用自己选择的名称; 记得用你的新接口返回一个参数化的List(例如List<SmallProject>)。

使用本机查询:

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

public static final String FIND_PROJECTS = "select ac_year_id,ac_year from tbl_au_academic_year where ac_year_id=?1";

    @Query(value = FIND_PROJECTS, nativeQuery = true)
    public  List<Object[]> findByAcYearId(Integer ac_year_id);

这对我很有用

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;
}