JPA& RequestFactory：外键的持久化

发布于 2024-11-18 08:35:53 字数 1615 浏览 4 评论 0原文

在我的 JPA 实体类中，我有：

@Entity
public class Booking {
@Id
@SequenceGenerator(name = "BOOKING", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "BOOKING")
private Integer id;

@ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
@JoinColumn(nullable = false)
private User user;

// ...
}

在我的 GWT 视图中，我有：

    MainRequestFactory.BookingRequest bookingRequest = reqFactory.bookingRequest();
    BookingProxy bookingProxy = bookingRequest.create(BookingProxy.class);
    UserProxy userProxy = bookingRequest.create(UserProxy.class);

    userProxy.setId(12);

    bookingProxy.setUser(userProxy);

    Request<Void> persistRequest = bookingRequest.persist().using(bookingProxy);
    persistRequest.fire(new Receiver<Void>() {
        @Override
        public void onSuccess(Void response) {
            GWT.log("persisted");
        }
    });

/////////////////////

Users 和 Bookings< /code> 没有用户约束的持久性工作正常。但通过上面的代码，我想要/必须将 ID 为“12”的用户分配给新创建的预订。但我无法将新的预订分配给已经存在的用户 ID 12。我收到以下错误：

[EL警告]：2011-07-05 18:48:45.464--UnitOfWork(267787945)--异常 [EclipseLink-4002]（Eclipse 持久性服务 - 2.3.0.v20110604-r9504): org.eclipse.persistence.exceptions.DatabaseException 内部异常： org.firebirdsql.jdbc.FBSQLException： GDS 例外。 335544665.违规主键或唯一键约束表“USERS2”上的“INTEG_388”

这是因为 ID 为“12”的用户已存在，并且 JPA 希望创建具有相同 ID 的新用户，而不是仅使用用户 ID“12”作为外键创建新预订。

我如何告诉 RequestFactory 不要创建新用户，而只是将现有用户的用户 ID 分配给新的预订条目？

原文

In my JPA Entity class I have:

@Entity
public class Booking {
@Id
@SequenceGenerator(name = "BOOKING", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "BOOKING")
private Integer id;

@ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
@JoinColumn(nullable = false)
private User user;

// ...
}

In my GWT View I have:

    MainRequestFactory.BookingRequest bookingRequest = reqFactory.bookingRequest();
    BookingProxy bookingProxy = bookingRequest.create(BookingProxy.class);
    UserProxy userProxy = bookingRequest.create(UserProxy.class);

    userProxy.setId(12);

    bookingProxy.setUser(userProxy);

    Request<Void> persistRequest = bookingRequest.persist().using(bookingProxy);
    persistRequest.fire(new Receiver<Void>() {
        @Override
        public void onSuccess(Void response) {
            GWT.log("persisted");
        }
    });

/////////////////////

Users and Bookings persistence without the User Constraint works fine. But with the above code I want/have to assign the user with the ID "12" to a newly created booking. But I cannot assign a new booking to an already existing user ID 12. I get the following error:

[EL Warning]: 2011-07-05
18:48:45.464--UnitOfWork(267787945)--Exception
[EclipseLink-4002] (Eclipse
Persistence Services -
2.3.0.v20110604-r9504): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception:
org.firebirdsql.jdbc.FBSQLException:
GDS Exception. 335544665. violation of
PRIMARY or UNIQUE KEY constraint
"INTEG_388" on table "USERS2"

It's because an user with the ID "12" already exists and JPA wants to create a new user with the same ID instead of just creating a new booking with the user ID "12" as its foreign key.

How can I tell RequestFactory NOT to create a new user but instead just to assign the user ID of an already existing user to a new booking entry?

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