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休眠一对一映射注释插入异常(在插入跨国表时,主表数据不应更改)

[英]Hibernate one to one mapping annotation insert exception (While inserting into transnational table, master table data should not change)

Oracle表DDL:

CREATE TABLE M_SERVICE
(
   SERVICE_ID    NUMBER (10) PRIMARY KEY,
   SERVICE_NM    VARCHAR2 (255),
   ACTIVE_SW     CHAR (1),
   LST_UPDT_DT   DATE,
   LST_UPDT_BY   VARCHAR2 (32)
)

CREATE TABLE T_JOB
(
   JOB_ID          NUMBER PRIMARY KEY,
   JOB_NM          VARCHAR2 (32),
   JOB_DESC        VARCHAR2 (2000),
   SERVICE_ID      NUMBER,
   DUE_DT          DATE,
   LST_UPDT_DT     DATE,
   LST_UPDT_BY     VARCHAR2 (32),
   CONSTRAINT T_JOB_FK1 FOREIGN KEY
      (SERVICE_ID)
       REFERENCES M_SERVICE (SERVICE_ID)
)

M_SERVICE是一个主表。 T_JOB是一个事务表。 我的要求是,当我尝试在T_job表中插入时,不应在M_service表中插入/更新(所有service_Id在M_SERVICE中都可用)。 但是在选择时,我需要两个表数据。

实体:

@Entity
@Table(name = "M_SERVICE")
public class ServiceVO implements Serializable {
    private static final long serialVersionUID = -2684205897352720653L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "SEQ_SERVICE_ID")
    @SequenceGenerator(name = "SEQ_SERVICE_ID", sequenceName = "SEQ_SERVICE_ID")
    @Column(name = "SERVICE_ID")
    private Integer serviceId;

    @Column(name = "SERVICE_NM")
    private String serviceName;

    @Column(name = "ACTIVE_SW")
    private char activeSwitch;  

    @Column(name = "LST_UPDT_DT")
    @Temporal(TemporalType.DATE)
    private Date lastUpdatedDt;

    @Column(name = "LST_UPDT_BY")
    private String lastUpdatedBy;
 //Getter & Setters.
}

@Entity
@Table(name = "T_JOB")
public class JobVO implements Serializable {
    private static final long serialVersionUID = 7167763557817486917L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "SEQ_JOB_ID")
    @SequenceGenerator(name = "SEQ_JOB_ID", sequenceName = "SEQ_JOB_ID")
    @Column(name = "JOB_ID")
    private Integer jobId;

    @Column(name = "JOB_NM", nullable = false)
    private String jobName;

    @Column(name = "JOB_DESC")
    private String jobDesc;

    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "SERVICE_ID", referencedColumnName = "SERVICE_ID", insertable = false, updatable = false)
    private ServiceVO service;

    @Column(name = "DUE_DT", nullable = false)
    @Temporal(TemporalType.DATE)
    private Date dueDate;

    @Column(name = "LST_UPDT_DT", nullable = false)
    @Temporal(TemporalType.DATE)
    private Date lastUpdatedDt;

    @Column(name = "LST_UPDT_BY", nullable = false)
    private String lastUpdatedBy;
//Getter & Setters.
}

如果我打电话要求根据工作ID表示特定的工作详细信息,那么我可以像下面期望的那样获得信息,

this.sessionFactory.getCurrentSession().get(JobVO.class, id);

JobVO [jobId = 1,jobName = job name,jobDesc = desc,service = ServiceVO [serviceId = 2,serviceName = Tax Audit,activeSwitch = Y,lastUpdatedDt = 2013-08-31,lastUpdatedBy = MAHESH],dueDate = 2013-09 -01,lastUpdatedDt = 2013-09-01,lastUpdatedBy = mahesh]

但是,如果我尝试插入工作方式,

            JobVO jobVO = new JobVO();
        jobVO.setAuditorId(1);                  
        jobVO.setDueDate(new Date());
        jobVO.setJobDesc("VAT1");
        jobVO.setJobName("Account");
        jobVO.setLastUpdatedBy("Mahesh");
        jobVO.setLastUpdatedDt(new Date());
        ServiceVO service=new ServiceVO();
        service.setServiceId(getClientId());
        jobVO.setService(service);
            this.sessionFactory.getCurrentSession().persist(jobVO);

出现如下所示的错误,org.springframework.orm.hibernate4.HibernateOptimisticLockingFailureException:批处理更新从更新[0]返回了意外的行数; 实际行数:0; 预期:1; 嵌套的异常是org.hibernate.StaleStateException:批处理更新从更新[0]返回了意外的行数; 实际行数:0; 预期:org.springframework.orm.hibernate4.HibernateTransactionManager.convertHibernateAccessException(HibernateTransactionManager.java:680)处org.springframework.orm.hibernate4.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:181)为1

请帮助我解决此问题。

发生错误是因为有2个事务在同一记录上工作。 如果一个记录被2个事务读取,并且如果该记录首先被一个事务保存,则在第二个事务中将抛出乐观锁定异常,因为系统假定其他人都不会修改该记录。

您是否在批处理模式下使用多线程?

有不同的解决方案:

  1. 确保在任何给定时刻仅一次事务处理将要更新的记录
  2. 重试事务:确保重试之间不保留任何非事务状态。

您具有单向关系。 您必须添加ServiceVO类:

@OneToOne(fetch = FetchType.LAZY, mappedBy = "service")
private JobVO job;

这段代码创建了双向关系,我认为它可以解决您的问题。

请尝试这个。

    ServiceVO service=new ServiceVO();
    service.setServiceId(getClientId());
    this.sessionFactory.getCurrentSession().saveOrUpdate(service);

    JobVO jobVO = new JobVO();
    jobVO.setAuditorId(1);      
    jobVO.setClientId(getClientId());       
    jobVO.setDueDate(new Date());
    jobVO.setJobDesc("VAT1");
    jobVO.setJobName("Account");
    jobVO.setLastUpdatedBy("Mahesh");
    jobVO.setLastUpdatedDt(new Date());
    jobVO.setService(service);
    this.sessionFactory.getCurrentSession().saveOrUpdate(jobVO);

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