Jpa datanucleus EntityManager合并正在插入新行而不是更新

CustomerSurvey.java

 public class CustomerSurvey
    implements Serializable
 {

/**
 * 
 */
private static final   long  serialVersionUID = 1L;

@Id @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "CUSTOMER_SURVEY_SEQUENCE") @Column(name = "SURVEYID", nullable = false) private String surveyId;

@ManyToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL) @JoinColumn(name = "CUSTNO", referencedColumnName = "CustNO") private Customer               customer;
@Column(name = "AVGRATINGS") private double                                                                                                                 avgRatings;

@Column(name = "COMMENTS") private String                                                                                                                   comments;

@Column(name = "SENTON") private Date                                                                                                                       sentOn;

@Column(name = "RESPONDEDON") private Date                                                                                                                  respondedOn;

@Column(name = "RESPONDED") int                                                                                                                             responded;
@Basic(fetch = FetchType.EAGER) @OneToMany(mappedBy = "customerSurvey", cascade = CascadeType.ALL) private Set<SurveyResponse>                              responses;

@XmlTransient @OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.DETACH) @JoinColumn(name = "SRNO") private ServiceRequest                             serviceRequest;

@Column(name = "ATTACHMENT") @Lob @Basic(fetch = FetchType.EAGER) private byte[]                                                                            attachment;
--- getters and setters---

 public int hashCode()
     {
    return new HashCodeBuilder(17,31)
    .append(this.surveyId)
 //        .append(this.avgRatings)
 //        .append(this.getResponded())
    .toHashCode();

}

@Override
public boolean equals(Object obj)
{

    return new EqualsBuilder().
            append("SurveyId",this.surveyId)
//                .append("responded", this.getResponded())
//                .append("AvgRatings", this.getAvgRatings())
            .isEquals();
}

它不更新，而是插入新行。 在使用hashcode builder和equals builder后，我因以下检查而得到false： System.out.println（“：”+survey.equals（dbSurvey））的测试；给假 系统输出println（survey.equals（cs））的测试； 上面给出了错误的答案

---

请注意，根据JPA规范，“瞬态”和“分离”是不同的状态，在merge（）中具有不同的行为。瞬态将创建一个新对象。已与现有的更新分离。在修改之前获取分离的对象，或者阅读

上的“分离”部分，当传递到“合并”时，对象处于什么状态？这对merge的功能至关重要。当我检查状态时，NucleusJPAHelper.getObjectState（obj）也会告诉您：我看到obj状态：瞬态。我正在使用远程客户端进行呼叫（REST呼叫）。当我从容器内部发出相同的调用时，我看到状态是分离的。谢谢你引导我找到它的状态。。。有没有关于如何处理这个差距的建议？我已经编辑了我的问题。。。我正在使用hashcode builder和equal builder，其中对象比较现在返回false。这有助于处理带有键的瞬态对象。设置datanucleus.allowAttachOfTransient=true是在实体管理器工厂（persistenct）级别完成的。它还影响那些真正瞬态的对象（key为null，实体尚未在db中创建）。它也尝试合并这些对象。如果我只在transient+有一个现有密钥的情况下寻找合并，会怎么样？目前我使用的是：CustomerSurvey refSurvey=em.find（CustomerSurvey.class，survey.getSurveyId（））；如果（（NucleusJPAHelper.getObjectState（survey）.equals（“transient”）&&（refSurvey！=null））{…持久性属性的描述非常清楚，显然，在调用merge（）时，任何没有id的临时对象都将被持久化没有附上，事实上，我们所有的测试都证明了这一事实，并且回答了您发布的问题

    if ( this.entityManagerFactory == null )
    {
        this.entityManagerFactory = this.getEntityManagerFactory();
    }

    EntityManager em = this.entityManagerFactory.createEntityManager();
    EntityTransaction tx = em.getTransaction();

    for (CustomerSurvey survey : surveys)
    {
        tx.begin();
        survey.setRespondedOn(Calendar.getInstance().getTime());
        survey.setResponded(BooleanResponse.YES.getCode());
        survey.setCustomer(em.find(Customer.class, survey.getCustomer().getCustNo().trim()));
        survey.setComments("Survey Response submitted");
        for (SurveyResponse response : survey.getResponses())
        {
            response.setQuestionair(em.find(Questionair.class, response.getQuestionair().getqNo()));
        }

        CustomerSurvey cs = em.find(CustomerSurvey.class, survey.getSurveyId());
        System.out.println(survey.equals(cs));
        Query query1 = em.createQuery("SELECT  from CustomerSurvey customerSurvey where customerSurvey.surveyId = :id")
                .setParameter("id", survey.getSurveyId());
        CustomerSurvey dbSurvey = (CustomerSurvey) query1.getResultList().get(0);
        System.out.println(":" + survey.equals(dbSurvey));

        if(survey.getServiceRequest() == null) {
            cs = em.find(CustomerSurvey.class, survey.getSurveyId());
            survey.setServiceRequest(cs.getServiceRequest());
        }
        CustomerSurvey mSurvey = em.merge(survey);

        String surveyId = mSurvey.getSurveyId();
        Query query = em.createNamedQuery("surveyResponse.getAverageRatings");
        logger.trace(query.toString());
        query.setParameter("surveyId", surveyId);
        logger.trace(query.toString());

        Double avgRating = (Double) query.getSingleResult();
        if ( avgRating == null )
        {
            avgRating = new Double(0);
        }
        BigDecimal bd = BigDecimal.valueOf(avgRating);
        bd = bd.setScale(2, RoundingMode.HALF_UP);
        avgRating = bd.doubleValue();
        mSurvey.setAvgRatings(avgRating);
        em.flush();
        tx.commit();
   }
    em.clear();
    em.close();
}