JPA - ManyToOne - JoinTable - 无法更新字段答案

【问题标题】：JPA - ManyToOne - JoinTable - cannot update a fieldJPA - ManyToOne - JoinTable - 无法更新字段
【发布时间】：2017-01-15 18:30:01
【问题描述】：

我有以下情况：

AnagraficaIscritti.java

@Entity
@Table(name="ANAGRAFICA_ISCRITTI")
public class AnagraficaIscritti implements Serializable {
        private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="ID_EMAIL",insertable=false, updatable=false)
    private long idEmail;

    private String email;

    //bi-directional many-to-many association to Newsletter 
    @ManyToMany(mappedBy="anagraficaIscrittis")
    private List<Newsletter> newsletters;

    //bi-directional many-to-one association to IscrizioniEmail
    @OneToMany(mappedBy="anagraficaIscritti")
    private List<IscrizioniEmail> iscrizioniEmails;

Newsletter.java

@Entity
@Table(name="NEWSLETTER")
public class Newsletter implements Serializable {
        private static final long serialVersionUID = 1L;

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        @Column(name="ID_NEWSLETTER")
        private long idNewsletter;

        //bi-directional many-to-many association to AnagraficaIscritti
        @ManyToMany
        @JoinTable(
            name="ISCRIZIONI_EMAIL"
            , joinColumns={
                @JoinColumn(name="ID_NEWSLETTER")
                }
            , inverseJoinColumns={
                @JoinColumn(name="ID_EMAIL")
                }
            )
        private List<AnagraficaIscritti> anagraficaIscrittis;

        //bi-directional many-to-one association to IscrizioniEmail
        @OneToMany(mappedBy="newsletter")
        private List<IscrizioniEmail> iscrizioniEmails;

IscrizioniEmail.java

@Entity
@Table(name="iscrizioni_email")
public class IscrizioniEmail implements Serializable {
        private static final long serialVersionUID = 1L;


        @EmbeddedId
        private IscrizioniEmailPK id;

        @Temporal(TemporalType.TIMESTAMP)
        @Column(name="SUBSCRIPTION_DATE")
        private Date subscriptionDate;

        //bi-directional many-to-one association to AnagraficaIscritti
        @ManyToOne
        @JoinColumn(name="ID_EMAIL",insertable = false, updatable = false)
        private AnagraficaIscritti anagraficaIscritti;

        //bi-directional many-to-one association to Newsletter
        @ManyToOne
        @JoinColumn(name="ID_NEWSLETTER", insertable = false, updatable = false)
        private Newsletter newsletter;

IscrizioniEmailPK.java

@Embeddable
public class IscrizioniEmailPK implements Serializable {
        //default serial version id, required for serializable classes.
        private static final long serialVersionUID = 1L;

        @Column(name="ID_EMAIL", insertable=false, updatable=false)
        private long idEmail;

        @Column(name="ID_NEWSLETTER", insertable=false, updatable=false)
        private long idNewsletter;

当我尝试创建 Object AnagraficaIscritti 和 Object Newsletter 时，也会在 IscrizioniEmail 中自动插入一条记录。例如我创建：

 ANAGRAFICA_ISCRITTI
 ID_EMAIL           EMAIL
  1           john.doe@gmail.com

 NEWSLETTER
 ID_NEWSLETTER      NEWSLETTER
      1               sport

 ISCRIZIONI_EMAIL
  ID_EMAIL   ID_NEWSLETTER   SUBSCRIPTION_DATE
     1             1               NULL

但是更新 SUBSCRIPTION_DATE（使用当前时间戳）的正确方法是什么，它是空的？如果我尝试创建一个 Object IscrizioniEmail，我会遇到重复的键错误等等。

【问题讨论】：

您是否尝试过创建@Entity 类IscrizioniEmail？
我不明白。实体类 IscrizioniEmail 存在。
谢谢，定义默认值有效：SUBSCRIPTION_DATEDATETIME DEFAULT NOW() 我认为 JPA 有时非常复杂。

标签： java jpa jointable

【解决方案1】：

您可以在ISCRIZIONI_EMAIL 表中为SUBSCRIPTION_DATE 设置默认值，该值将是当前日期。这样您就不需要从 Java 发送它。

如果您需要从 Java 发送该值，则可以在插入 Newsletter 和 AnagraficaIscritti 表之后使用 IscrizioniEmail 实体类执行更新查询。

【讨论】：