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按方法参数的条件映射

mapstruct mapping

José Sierra · 技术社区 · 2 年前

我有这个jpa:

public class CandidateRecommendationJpa extends AuditingEntityJpa {
  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "ID_CANDIDATO_RECOMENDACION", nullable = false)
  private Integer id;

  @ManyToOne
  @JoinColumn(name = "ID_CANDIDATO_EMISOR", nullable = false)
  private CandidateJpa candidateSender;

  @ManyToOne
  @JoinColumn(name = "ID_CANDIDATO_RECEPTOR", nullable = false)
  private CandidateJpa candidateReceiver;
}

我必须映射这个实体:

@Getter
@Setter
@AllArgsConstructor
@NoArgsConstructor
@Builder
@EqualsAndHashCode
public class CandidateRecommendation {

  private Integer id;   

  private Candidate candidate;

}

映射器:

@Mapper(componentModel = "spring")
public interface CandidateRecommendationJpaMapper {

  CandidateRecommendationJpaMapper INSTANCE = Mappers.getMapper(CandidateRecommendationJpaMapper.class);

  @Mapping(target = "candidate.id", source = "candidateSender.id")
  @Mapping(target = "candidate.name", source = "candidateSender.name")
  @Mapping(target = "candidate.login", source = "candidateSender.employee.login")
  CandidateRecommendation toModel(CandidateRecommendationJpa jpa);

  List<CandidateRecommendation> toModels(List<CandidateRecommendationJpa> jpa);

}

问题

我必须指定jpa-atribute来映射,有没有办法用param方法映射jpa属性?例如,在这种情况下,我使用candidateSender进行映射,但是,如何使用candiateReceiver进行映射?

  @Mapping(target = "candidate.id", source = "candidateReceiver.id")
  @Mapping(target = "candidate.name", source = "candidateReceiver.name")
  @Mapping(target = "candidate.login", source = "candidateReceiver.employee.login")
  CandidateRecommendation toSenderModel(CandidateRecommendationJpa jpa);

  List<CandidateRecommendation> toSenderModels(List<CandidateRecommendationJpa> jpa);

这显示以下错误:

Ambiguous mapping methods found for mapping collection element to CandidateRecommendation: CandidateRecommendation toModel(CandidateRecommendationJpa jpa), CandidateRecommendation toSenderModel(CandidateRecommendationJpa jpa)

0 回复 | 直到 2 年前

Paul Marcelin Bejan 2 年前

问题在于:

List<CandidateRecommendation> toSenderModels(List<CandidateRecommendationJpa> jpa);

不知道使用哪种映射方法,因为有两种方法:

to模型
至发送方模型

因此,要解决此问题,您需要执行以下操作:

@Mapper(componentModel = "spring")
public interface CandidateRecommendationJpaMapper {

    CandidateRecommendationJpaMapper INSTANCE = Mappers.getMapper(CandidateRecommendationJpaMapper.class);

    @Named("toModel")
    @Mapping(target = "candidate.id", source = "candidateSender.id")
    @Mapping(target = "candidate.name", source = "candidateSender.name")
    @Mapping(target = "candidate.login", source = "candidateSender.employee.login")
    CandidateRecommendation toModel(CandidateRecommendationJpa jpa);

    @Named("toSenderModel")
    @Mapping(target = "candidate.id", source = "candidateReceiver.id")
    @Mapping(target = "candidate.name", source = "candidateReceiver.name")
    @Mapping(target = "candidate.login", source = "candidateReceiver.employee.login")
    CandidateRecommendation toSenderModel(CandidateRecommendationJpa jpa);

    @IterableMapping(qualifiedByName = "toModel")
    List<CandidateRecommendation> toModels(List<CandidateRecommendationJpa> jpa);

    @IterableMapping(qualifiedByName = "toSenderModel")
    List<CandidateRecommendation> toSenderModels(List<CandidateRecommendationJpa> jpa);

}

正如您所看到的,我添加了@Named和@IterableMapping。