如何使用JPA在特定实体组中创建对象？ （Google AppEngine Java）

Question

我想在JPE中使用GAE-J中的事务。

没有JPA它应该是：

Entity child= new Entity("Child", "ParentKey"); 

，但如何使用JPA办呢？

@Entity 
public class Parent{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Key id; 
    private String text; 
} 

@Entity 
public class Child{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Key id; 
    private Parent parent; 
    private String text; 
} 

尝试...

Parent parent = new Parent(); 
em.persist(parent); 
em.refresh(parent); 

Child child = new Child(); 
child.setParent(parent); 
em.persist(child); 

这不起作用：

org.datanucleus.store.appengine.DatastoreRelationFieldManager$ChildWithoutParentException: 
Detected attempt to establish Child(130007) as the parent of Parent(132001) but the entity identified by Child(132001) has already been persisted without a parent. A parent cannot be established or changed once an object has been persisted. 

这听起来有点后到前... 我是一个傻瓜？还是有一个简单的方法？

谢谢！

Answer 1

凯...我的第一次尝试只是一个小错误。

这应该工作：

@Entity 
public class Parent{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Key id; 
    private String text; 
    @OneToMany(targetEntity=Child.class, mappedBy="parent", fetch=FetchType.LAZY) 
    private Set<Child> children = new HashSet<Child>(); 
} 

@Entity 
public class Child{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Key id; 
    @ManyToOne(fetch=FetchType.LAZY, targetEntity=Parent.class) 
    private Parent parent; 
    private String text; 
}