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我正在使用Spring的JPA使用namedquery來獲取其他api。命名查詢是在我的實體類實現:IllegalArgumentException:使用Spring的NamedQuery JPA
@Entity
@Table(name="SPECIMEN_TB")
@NamedQueries({
@NamedQuery(name="SpecimenTb.findBySpecimenNo", query="select s from SpecimenTb s where s.specimenNo = :specimenNo"),
})
public class SpecimenTb implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SPECIMEN_TB_ROWID_GENERATOR")
@Column(name="ROW_ID")
private long rowId;
@Column(name="SPECIMEN_NO", unique = true)
private String specimenNo;
我的控制器看起來是這樣的:
@RestController
public class RistoreController {
@Autowired
private RistoreService ristoreService;
@RequestMapping(
value = "/ristore/foundation/{specno}",
method = RequestMethod.GET,
produces = "application/json")
public ResponseEntity<SpecimenTb> getFmSpecimen(@PathVariable("specno") String specno) {
List<SpecimenTb> specimens = ristoreService.findBySpecimenNo(specno);
if (specimens == null) {
return new ResponseEntity<SpecimenTb>(HttpStatus.NOT_FOUND);
}
return new ResponseEntity<SpecimenTb>(specimens.get(0), HttpStatus.OK);
}
我有一個服務Bean上調用JPA庫findBySpecimenNo方法。
@Service
public class RistoreServiceBean implements RistoreService {
@Autowired
private SpecimenRepository specimenRepository;
@Override
public List<SpecimenTb> findAll() {
List<SpecimenTb> specimens = specimenRepository.findAll();
return specimens;
}
@Override
public List<SpecimenTb> findBySpecimenNo(String specimenNo) {
List<SpecimenTb> specimens = specimenRepository.findBySpecimenNo(specimenNo);
return specimens;
}
當我開始春季啓動應用程序並輸入網址「http://localhost:8080/ristore/foundation/SKM1」,我得到了以下錯誤:
java.lang.IllegalArgumentException: Parameter with that position [1] did not exist
我做了什麼錯?
它現在正常工作,如果我想使命名參數工作,我應該在哪裏「在方法參數上添加註釋」? – ddd
我沒有看到您的存儲庫接口/類,但這是它將完成的地方。查看'@ Query'周圍的spring-data-jpa文檔 –