Question

我在我的应用程序中使用EJB 3和JPA 2，我有2个实体，有许多关系，并希望坚持它们但幸运的是经过数小时的搜索并尝试我无法做到，所以这个是我的代码：

@Entity
public class Destinataire implements Serializable {

    @Id
    @Basic(optional = false)
    @Size(min = 1, max = 50)
    @Column(name = "ADRESSE")
    private String adresse;
    @OneToMany(mappedBy = "destinataire")
    private Set<Envois> envoisSet;

//getters and setters
}

@Entity
public class Envois implements Serializable {

    @EmbeddedId
    protected EnvoisPK envoisPK = new EnvoisPK();
    @Size(max = 4)
    @Column(name = "TYPENVOIS")
    private String typenvois;
    @JoinColumn(name = "IDMAIL", referencedColumnName = "IDMAIL", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private MailAEnvoyer mailAEnvoyer;
    @JoinColumn(name = "ADRESSE", referencedColumnName = "ADRESSE", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private Destinataire destinataire;
}

@Embeddable
public class EnvoisPK implements Serializable {

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 255)
    @Column(name = "ADRESSE")
    private String adresse;
    @Basic(optional = false)
    @NotNull
    @Column(name = "IDMAIL")
    private BigDecimal idmail;
}

@Entity
@Table(name = "MAILAENVOYER")
public class MailAEnvoyer implements Serializable {

    private static final long serialVersionUID = 1L;
    // @Max(value=?)  @Min(value=?)//if you know range of your decimal fields consider using these annotations to enforce field validation
    @Id
    @Basic(optional = false)
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    @Column(name = "IDMAIL")
    private BigDecimal idmail;



    @JoinTable(name = "ENVOIS", joinColumns = {
        @JoinColumn(name = "IDMAIL", referencedColumnName = "IDMAIL")}, inverseJoinColumns = {
        @JoinColumn(name = "ADRESSE", referencedColumnName = "ADRESSE")})
    @ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.MERGE)
    private Set<Destinataire> destinataireSet;
    @OneToMany(mappedBy = "mailAEnvoyer", cascade = CascadeType.ALL)
    private Set<Envois> envoisSet;
}

这就是我的进展方式：

Envois e = new Envois("CCI");
Destinataire dest = new Destinataire("adress");
destinataireService.create(dest);
e.setDestinataire(dest);
currentMailAEnvoyer.getEnvoisSet().add(e);
mailService.save(currentMailAEnvoyer);

但我总是得到一个例外，EnvoisPK有adresse和idmail null，

Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLIntegrityConstraintViolationException: ORA-01400: cannot insert NULL into ("MAILING"."ENVOIS"."IDMAIL")
Error Code: 1400
Call: INSERT INTO ENVOIS (TYPENVOIS, IDMAIL, ADRESSE) VALUES (?, ?, ?)
    bind => [3 parameters bound]
Query: InsertObjectQuery(entities.Envois[ envoisPK=entities.EnvoisPK[ adresse=null, idmail=null ] ])

Answer 1

您可以在实体MailAEnvoyer中添加以下方法：

@PostPersist
public void postPersist() {
    for (Envois e : envoisSet) {
        e.getEnvoisPK().setIdmail(this.getIdmail());
    }

}

使用jpa映射多对多关系和Embeddable类

1 个答案: