通过方法参数进行条件映射答案

【问题标题】：Conditional mapping by method parameter通过方法参数进行条件映射
【发布时间】：2022-11-20 23:10:44
【问题描述】：

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

我必须指定要映射的 jpa 属性，有没有办法用参数方法映射 jpa 属性？例如，在这种情况下，我映射到候选发送者，但是我如何映射到候选接收者？

  @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)

【问题讨论】：

标签： java mapping mapstruct

【解决方案1】：

问题是：

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

不知道使用哪种映射方法，因为有两种：

到模型
toSenderModel

因此，要解决该问题，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);

}

如您所见，我添加了@Named 和@IterableMapping。

【讨论】：