按field Spring Boot查找

Question

@Entity
@Table(name="paciente")
public class Paciente {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private long id;

    @Column(name="rut_num")
    private int rut_num;

    @Column(name="nombre")
    private String nombre;

    @Column(name="a_pat")
    private String a_pat;

    @Column(name="a_mat")
    private String a_mat;

    //getters and setters

@Repository("pacienteRepository")
public interface PacienteJpaRepository extends JpaRepository<Paciente, Serializable> {

    @Query("SELECT nombre FROM paciente WHERE rut_num= :rut_num")
    Future<String> findByRut(@Param("rut_num") int rut_num);
}

public interface PacienteService {

    public abstract Paciente findByRut(int rut_num);
}

@Service("pacienteServiceImpl")
public class PacienteServiceImpl implements PacienteService {

    @Autowired
    @Qualifier("pacienteRepository")
    private PacienteJpaRepository pacienteRepository;

    @Override
    public Paciente findByRut(int rut_num) {
        return (Paciente) pacienteRepository.findByRut(rut_num);
    }

}

@Controller
public class PacienteController {

    @Autowired
    @Qualifier("pacienteServiceImpl")
    private PacienteService pacienteService;

    @RequestMapping(value = "/paciente/{rut_num}", method = RequestMethod.GET, headers = "Accept=application/json")
    public ResponseEntity<Paciente> findByRut(@PathVariable("rut_num") int rut_num){
        Paciente paciente = pacienteService.findByRut(rut_num);
        if (paciente == null) {
            return new ResponseEntity(HttpStatus.NOT_FOUND);
            // You many decide to return HttpStatus.NOT_FOUND
        }
        return new ResponseEntity<Paciente>(paciente, HttpStatus.OK);
    }
}

控制台错误：

创建名为“Paciente ServiceImpl”的bean时出错：通过字段“Paciente Repository”表示的依赖项不满足；嵌套异常为org.springframework.beans.factory.BeanCreationException：创建名为“Paciente Repository”的bean时出错：调用init方法失败；嵌套异常为java.lang.IllegalArgumentException：对方法public abstract java.util.concurrent.future com.calendar.repository.PacientEjparePository.findByRut(int)的查询验证失败！

Answer 1

我认为问题是您在查询中重新检索到的实体的名称是“Paciente”而不是“Paciente”，因此更改

@Query("SELECT nombre FROM paciente WHERE rut_num= :rut_num")
Future<String> findByRut(@Param("rut_num") int rut_num);

至

@Query("SELECT nombre FROM Paciente WHERE rut_num= :rut_num")
Future<String> findByRut(@Param("rut_num") int rut_num);

应该工作