休眠 - 如何用分离的孩子保存父母 [英] hibernate - how to save parent with detached child

问题描述

我正在从 UI 发送一个对象.将使用对现有子项的引用来创建此对象.

I am sending an object from UI. This object is going to be created with a reference to an existing child.

这是这种关系的简单说明.

This is simple illustration for this relation.

class ParentEntity {
    @Id
    Long id;

    @ManyToOne(fetch = FetchType.LAZY)
    private ChildEntity child;
}

class ChildEntity {
    @Id
    Long id;
}



ChildEntity child = new ChildEntity();
child.setId(1);
//parentEntity is created based on data sent from UI
parentEntity.setChild(child);

当我保存这个对象时，Hibernate 给了我"org.hibernate.TransientPropertyValueException:对象引用了一个未保存的瞬态实例".

When I save this object, Hibernate gives me " org.hibernate.TransientPropertyValueException: object references an unsaved transient instance ".

我不需要去救孩子，因为我根本不换孩子.只需要在父表中保存孩子的id.

I don't have to save a child as I do not change child at all. Just need to save child's id in parent's table.

我尝试使用了一些 CascadeType，但都没有奏效.

I've tried to use few CascadeType, but none of them worked.

推荐答案

只需为孩子使用代理:

parentEntity.setChild(entityManager.getReference(ChildEntity.class, childId));

这里的重点是使用 EntityManager.getReference:

获取一个实例，其状态可能会被延迟获取.

Get an instance, whose state may be lazily fetched.

Hibernate 将创建只包含 id 而不去数据库的代理.

Hibernate will create the proxy containing only the id without going to the database.

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