Question

我想通过以下示例实现的目的是避免发生的约束违规异常运行以下内容时：

Parent p = new Parent();
Set<Child> children = new HashSet<Child>();
Child c = new Child();
children.add(c);
p.setChildren(children);

entityManager.merge(p);
entityManager.merge(p);

如何配置JPA来实现此效果？以下是课程。

@Entity
@Table(name = "parent", indexes = {uniqueConstraints = { @UniqueConstraint(columnNames = { "uri" }) })
public class Parent implements Serializable {
  ...
  private String uri;
  private Set<Child> children = new HashSet<Child>();


  @Id
  @Column(name = "id", unique = true)
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "parent_seq")
  @SequenceGenerator(name = "parent_seq", sequenceName = "parent_seq", allocationSize = 1)
  public Long getId() {
    return id;
  }

  @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true)
  @JoinColumn(name = "parent_id")
  public Set<Child> getChildren() {
    return children;
  }

@Override
  public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Parent other = (Parent) obj;
        if (uri == null) {
            if (other.uri != null)
                return false;
        } else if (!uri.equals(other.uri))
            return false;
        return true;
    }


@Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((uri == null) ? 0 : uri.hashCode());
        return result;
    }



}

@Entity
@Table(name = "child", indexes = {uniqueConstraints = { @UniqueConstraint(columnNames = { "text" }) })
public class Child implements Serializable {

  private String text;

  @Id
  @Column(name = "id", unique = true)
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "child_seq")
  @SequenceGenerator(name = "child_seq", sequenceName = "child_seq", allocationSize = 1)
  public Long getId() {
    return id;
  }



  @Override
  public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Child other = (Child) obj;
        if (text == null) {
            if (other.text != null)
                return false;
        } else if (!text.equals(other.text))
            return false;
        return true;
    }


@Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((text == null) ? 0 : text.hashCode());
        return result;
    }
}

我使用postgres和hibernate作为JPA提供程序

合并时避免JPA和OneToMany关系中的ConstraintViolationException

0 个答案: