java - Hibernate - 使用组合键保留 @OneToOne

Question

我有以下“审核”表，可以托管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 Audit 类的父类(super class):

@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;
    }

}

此映射在从数据库检索实体及其审核信息时有效，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

我已经尝试了所有我能想到的或在网上找到的东西，最后总是归结为同样的问题:Hibernate 尝试插入我的 Audit 对象，其中 的值为空>item_id 和版本。

如果我手动将我的 entity 实例和版本设置为 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模糊的错误:

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，也无法更改 Hibernate 的版本(5.1.0，我知道更高版本中修复了一些错误，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 EntityAAudit.item 上的 @MapsId 注释。

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

JPA 2.2 spec 中讨论了派生身份(并附有示例)在第 2.4.1 节中。