时间戳

首页 > 解决方案 > 有没有办法使用 hibernate-envers 审计一个在其 EmbeddedId 中具有 @Embedded 的实体

java - 有没有办法使用 hibernate-envers 审计一个在其 EmbeddedId 中具有 @Embedded 的实体

问题描述

我有一个具有复合代码 BlocRecordId 的实体 BlocRecord，并且在其复合代码中有一个 @Embedded（关系代码 ManyToOne）指向另一个实体记录并想要审核实体 BlocRecord。

实体 BlocRecord

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@Table(name = "blocRecord")
@Access(value = AccessType.FIELD)
@Audited
public class BlocRecord {

    @EmbeddedId

    private BlocRecordId blocRecordId = new BlocRecordId();

    @ManyToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumns({
            @JoinColumn(name = "record_identifier_", referencedColumnName = "identifier_", unique = false, nullable = false),
            @JoinColumn(name = "record_recordType_", referencedColumnName = "recordType_", unique = false, nullable = false)})
    @MapsId("record")
    private Record record;

...

}

id 类 BlocRecordID

@Embeddable
public class BlocRecordId implements Serializable {


    @Embedded
    private RecordId record;

    @Column(name = "source_")
    String source ;

    @Column(name = "messageType_")
    String messageType ;

实体记录

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@Table(name = "records")
@Access(value = AccessType.FIELD)
@Audited
public class Record {

    @EmbeddedId
    private RecordId recordId = new RecordId();


    @OneToMany(targetEntity = BlocRecord.class, fetch = FetchType.LAZY, mappedBy = "record")
    private Set<BlocRecord> blocRecord = new java.util.HashSet<>();

...
}

实体 Record 的 idClass

@Embeddable
public class RecordId implements Serializable{

    @Column(name = "identifier_")
    String identifier ;

    @Column(name = "recordType_")
    String recordType ;

}

Hibernate-envers 在尝试生成可嵌入的 BlocRecordId 中的字段记录的元数据时失败，抛出流异常

org.hibernate.MappingException: Type not supported: org.hibernate.type.ComponentType
    at org.hibernate.envers.configuration.internal.metadata.IdMetadataGenerator.addIdProperties(IdMetadataGenerator.java:121)
    at org.hibernate.envers.configuration.internal.metadata.IdMetadataGenerator.addId(IdMetadataGenerator.java:230)
    at org.hibernate.envers.configuration.internal.metadata.AuditMetadataGenerator.generateFirstPass(AuditMetadataGenerator.java:642)
    at org.hibernate.envers.configuration.internal.EntitiesConfigurator.configure(EntitiesConfigurator.java:95)
    at org.hibernate.envers.boot.internal.EnversServiceImpl.doInitialize(EnversServiceImpl.java:154)
    at org.hibernate.envers.boot.internal.EnversServiceImpl.initialize(EnversServiceImpl.java:118)
    at org.hibernate.envers.boot.internal.AdditionalJaxbMappingProducerImpl.produceAdditionalMappings(AdditionalJaxbMappingProducerImpl.java:99)
    at org.hibernate.boot.model.process.spi.MetadataBuildingProcess.complete(MetadataBuildingProcess.java:288)
    at org.hibernate.boot.model.process.spi.MetadataBuildingProcess.build(MetadataBuildingProcess.java:83)
    at org.hibernate.boot.internal.MetadataBuilderImpl.build(MetadataBuilderImpl.java:417)
    at org.hibernate.boot.internal.MetadataBuilderImpl.build(MetadataBuilderImpl.java:86)
    at org.hibernate.boot.MetadataSources.buildMetadata(MetadataSources.java:179)

你知道如何解决这个问题吗？

谢谢

标签： javahibernatehibernate-envers

解决方案

目前，当我们像您的示例所示那样映射标识符列时，Envers 不支持将可嵌入对象嵌套在可嵌入对象中的想法。Envers 目前支持的唯一有效映射是嵌入对象中的属性是 a@ManyToOne还是@Basic类型。

您可以解决这个问题，但它需要更明确一些，而不是使用RecordId. 我的意思是重写BlocRecordId为以下内容：

@Embeddable
public class BlocRecordId implements Serializable {
  @Column(name = "identifier_")
  String identifier;

  @Column(name = "recordType_")
  String recordType;

  @Column(name = "source_")
  String source;

  @Column(name = "messageType_")
  String messageType;

  @Transient
  private RecordId recordId;

  /** Helper method to assign the values from an existing RecordId */
  public void setRecordId(RecordId recordId) {
    this.identifier = recordId.getIdentifier();
    this.recordType = recordId.getRecordType();
  }

  /** Helper method to get the RecordId, caching it to avoid multiple allocations */
  public RecordId getRecordId() {
    if ( recordId == null ) {
      this.recordId = new RecordId( identifier, recordType );
    }
    return this.recordId;
  }
}

我同意这不太理想，但它至少可以解决当前代码的限制。我已经添加并添加了HHH-13361作为一个开放问题来支持这一点。如果您愿意，欢迎您贡献，否则我将努力为 Envers 6.0 提供支持。

推荐阅读