Question

我具有以下“审核”表，该表可以承载有关我的架构的任何其他类的审核信息：

CREATE TABLE `audit` (
  `table_name` varchar(45) NOT NULL,
  `item_id` int(11) NOT NULL,
  `version` int(11) NOT NULL,
  `updated_at` datetime NOT NULL,
  `updated_by` varchar(25) NOT NULL,
  `comment` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`table_name`,`item_id`,`version`)
)

然后，我的架构中有不同的JPA实体，如下所示：

@Entity(name = "EntityA")
@Table(name = "entity_a")
public class EntityA {
    @Id
    @GeneratedValue
    private Long id;

    private Long version;

    // Other fields

    @OneToOne(mappedBy = "id.item", targetEntity = EntityAAudit.class, fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    private EntityAAudit audit;
}

同时，我有一个抽象类Audit，它是多个特定于实体的审计类的超类：

@MappedSuperclass
@Table(name = "audit")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "table_name", discriminatorType = DiscriminatorType.STRING)
@DiscriminatorOptions(insert = true, force = true)
public abstract class AuditHistory {

    // Some audit fields like the date and the author of the modification

}

@Entity(name = "EntityAAudit")
@Table(name = "audit")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorValue("entity_a")
@DiscriminatorOptions(insert = true, force = true)
public class EntityAAudit extends Audit {

    @EmbeddedId
    @JsonUnwrapped
    private AuditedId id;

    @Embeddable
    public static class AuditedId implements Serializable {
        @OneToOne
        @JoinColumn(name = "item_id", nullable = false)
        private EntityA item;

        @Column(name = "version", nullable = false)
        private Long version;
    }

}

当从数据库中检索实体及其审核信息时，此映射有效，但在插入具有相应审核信息的新实体时无效：

EntityA entity = new EntityA();
entity.setVersion(/* ... */);
// Setting the other basic fields of the entity

EntityAAudit audit = new EntityAAudit();
// Setting the basic fields of the audit

entity.setAudit(audit);

entity = em.merge(entity);

我最终遇到以下异常：

org.hibernate.id.IdentifierGenerationException: null id generated for:class EntityAAudit

我实际上尝试了所有我能想到的或可以在网上找到的所有内容，最后总是归结为同一问题：Hibernate尝试为我的Audit对象插入item_id的空值和version。

如果我手动设置entity实例和版本作为audit对象的ID，如下所示：

EntityA entity = new EntityA();
entity.setVersion(/* ... */);
// Setting the other basic fields of the entity

EntityAAudit audit = new EntityAAudit();
// Setting the basic fields of the audit
audit.setId(new EntityAAudit.AuditedId());
audit.getId().setItem(entity);
audit.getId().setVersion(entity.getVersion());

entity.setAudit(audit);

entity = em.merge(entity);

然后我在这里遇到了更加晦涩的错误：

Caused by: java.lang.NullPointerException
    at org.hibernate.type.descriptor.java.AbstractTypeDescriptor.extractHashCode(AbstractTypeDescriptor.java:65)
    at org.hibernate.type.AbstractStandardBasicType.getHashCode(AbstractStandardBasicType.java:185)
    at org.hibernate.type.AbstractStandardBasicType.getHashCode(AbstractStandardBasicType.java:189)
    at org.hibernate.type.EntityType.getHashCode(EntityType.java:348)

请注意，我无法更改数据库的结构或Hibernate的版本（5.1.0，我知道某些错误已在以后的版本中修复，可以解决我的问题...）。

非常感谢：）

Answer 1

您可以尝试“派生身份”映射：

@Entity(name = "EntityAAudit")
@Table(name = "audit")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorValue("entity_a")
@DiscriminatorOptions(insert = true, force = true)
public class EntityAAudit extends Audit {

    @EmbeddedId
    @JsonUnwrapped
    private AuditedId id;

    @OneToOne
    @JoinColumn(name = "item_id", nullable = false)
    @MapsId("entityAId") // maps entityAId attribute of embedded id
    private EntityA item;

    @Embeddable
    public static class AuditedId implements Serializable {
        private Long entityAId; // corresponds to PK type of EntityA

        @Column(name = "version", nullable = false)
        private Long version;
    }

}

请注意@MapsId上的EntityAAudit.item注释。

此外，您将需要显式设置EntityAAudit.item和AuditedId.version。 JPA不会神奇地为您确定和设置任何循环引用。

在JPA 2.2 spec的第2.4.1节中讨论了派生的身份（带有示例）。

Hibernate-使用复合键坚持@OneToOne

1 个答案: