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提交实体时出现以下错误。根据错误，我必须为实体分配一个 id，但我希望它必须由 JPA 本身处理。

Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.id.IdentifierGenerationException: ids for this class must be manually assigned before calling save(): alarm.ServiceAlarmConfEntity
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1214)
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1147)
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1153)
    at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:678)
    at alarm.Test.main(Test.java:32)
Caused by: org.hibernate.id.IdentifierGenerationException: ids for this class must be manually assigned before calling save(): alarm.ServiceAlarmConfEntity
    at org.hibernate.id.Assigned.generate(Assigned.java:53)
    at org.hibernate.event.def.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:121)
    at org.hibernate.ejb.event.EJB3PersistEventListener.saveWithGeneratedId(EJB3PersistEventListener.java:69)

我有两个具有一对一关系的实体。

@Entity(name ="services")
@Table(name = "services")
public class Service implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Column(name = "label")
    private String label;
    @Column(name = "schemapath")
    private String schemapath;
    @Column(name = "customizedPath")
    private String customizedPath;
    @Column(name = "birimId")
    private Integer birimId;   
    @Column(name = "alarmthresholdXML")
    private String alarmthresholdXML;
    @OneToOne(cascade = CascadeType.ALL, mappedBy = "service", fetch = FetchType.LAZY)
    private ServiceAlarmConfEntity serviceAlarmConfEntity;

@Entity(name ="service_alarm_conf")
@Table(name = "service_alarm_conf")
public class ServiceAlarmConfEntity implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @Column(name = "service_id")
    private Integer serviceId;
    @Lob
    @Column(name = "alarmthresholdXML")
    private String alarmthresholdXML;
    @Column(name = "alarmCheckEnable")
    private Integer alarmCheckEnable;
    @Column(name = "alarmDataTransferTimeInSeconds")
    private Integer alarmDataTransferTimeInSeconds;
    @JoinColumn(name = "service_id", referencedColumnName = "id", insertable = false, updatable = false)
    @OneToOne(optional = false, fetch = FetchType.LAZY)
    private Service service;

主类：

package alarm;


import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;

public class Test {

public static void main(String[] args) {
    EntityManagerFactory emf = Persistence.createEntityManagerFactory("alarm");

    EntityManager em = emf.createEntityManager();
    EntityTransaction tx = em.getTransaction();
    Service service = new Service();
    service.setLabel("Test1");
    service.setBirimId(16);
    service.setSchemapath("test1");
    service.setCustomizedPath("Test1");


    ServiceAlarmConfEntity serviceAlarmConfEntity = new ServiceAlarmConfEntity();
    serviceAlarmConfEntity.setAlarmCheckEnable(1);
    serviceAlarmConfEntity.setAlarmthresholdXML("alarmThresholdXML");
    serviceAlarmConfEntity.setService(service);


    service.setServiceAlarmConfEntity(serviceAlarmConfEntity);

    tx.begin();
    em.persist(service);        
    tx.commit();
    em.close();
}

}

更新：

ServiceAlarmConfEntity 配置如下。它有效。

   @MapsId
   @JoinColumn(name = "service_id", referencedColumnName = "id")
   @OneToOne(optional = false, fetch = FetchType.LAZY)
   private Service service;
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1 回答 1

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只有使用 注释的 ID 才会自动生成@GeneratedValue。但事实并非如此。

编辑：

如文档中所述，让子实体共享其父实体 ID 的标准和推荐方法如下：

public class ServiceAlarmConfEntity {
    @Id
    private Integer serviceId;

    @MapsId 
    @OneToOne(optional = false, fetch = FetchType.LAZY)
    @JoinColumn(name = "service_id")
    private Service service;

    ...
}
于 2013-06-03T18:34:41.277 回答