Hibernate 保持一对多单向关系时外键为空 上下文

我有两个实体，映射如下：

@Entity
public class User {

@Id
private String mail;

@OneToMany(cascade={CascadeType.REMOVE} fetch=FetchType.EAGER)
private Set<Food> foods;

我的测试：

@Test   
public void testPersistUserWithFood() throws Exception{
  //I create and persist the user
  User user = new User ("@someone");
  Session sess = HibernateUtil.getSessionFactory().openSession();
  Transaction tx = sess.beginTransaction();
  sess.save(user);
  tx.commit();
  sess.close();      

  //Now I create de food and set the relation in object schema
  Food foo = new Food();
  user.getFoods.add(foo);

  //I persist the food in the same way than user, with session, transaction and so on.

  //Then I retrieve the user from DB
  sess = HibernateUtil.getSessionFactory().openSession();
  User userRetrieved =(User) sess.createCriteria(User.class)
                 .add(Restrictions.idEq("@someone")).uniqueResult();
  sess.close();

  //Finally assert
  assertEquals(user.getFoods(), userRetrieved.getFoods());

}

问题 测试失败，因为调用：userRetrieved.getFoods（）返回空列表。 调试后，我看到table Food中的id_User foring列始终为null。 如果关系是oneToMany双向的，则列id_User不为null，因为我在Food类中添加了User属性

谢谢你的帮助

注意：我使用的是Hibernate4.3.8、JUnit4和Java8

userRetrieved.getFoods() return a null list.

这意味着在添加食物时不会更新用户

...
...

  //Now I create de food and set the relation in object schema

  //I persist the food in the same way than user, with session, transaction and so on.
  Session sess = HibernateUtil.getSessionFactory().openSession();
  Transaction tx = sess.beginTransaction();

  Food foo = new Food();
  user.getFoods.add(foo);

  sess.update(user);    //are you updating User as well like this?
  sess.save(foo);
  tx.commit();
  sess.close();
...
...

---

另外，如果它不起作用，那么请发布完整的代码，显示您是如何持久化所有实体的。

您拥有

id\u User

列

null

，因为

id

等于

null

在新的

Food

对象中

Food foo = new Food();
user.getFoods().add(foo);

要解决此问题，您需要将一个新的

Foo

对象保存到Hibernate，并为其分配一个

id

。然后将保存的

Foo

添加到

foods

集合中。或者您可以使用

cascade=CascadeType.ALL

将新的子对象与父对象一起休眠保存

@OneToMany(cascade = CascadeType.ALL, fetch=FetchType.EAGER)
private Set<Food> foods;

@OneToMany（cascade=CascadeType.ALL，fetch=FetchType.EAGER）
私家菜；

谢谢大家，我解决了我的问题，代码如下：

一个实体 试验 总结：

保存用户

节约食物

更新用户

非常感谢

@OneToMany(cascade = CascadeType.ALL, fetch=FetchType.EAGER)
private Set<Food> foods;

@Entity
public class User {

@Id
private String mail;

@OneToMany(cascade={CascadeType.REMOVE} fetch=FetchType.EAGER)
@JoinColumn(name="id_User")
private Set<Food> foods;

@Entity
public class Food{

@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private int id;

 @Test   
 public void testPersistUserWithFood() throws Exception{

  //I create and persist the user
  User user = new User ("@someone");
  Session sess = HibernateUtil.getSessionFactory().openSession();
  Transaction tx = sess.beginTransaction();
  sess.save(user);
  tx.commit();
  sess.close();      

  //Now I create de food and set the relation in object schema
  Food foo = new Food();
  user.getFoods.add(foo);

  //I persist the food in the same way than user, with session, transaction and so on.

  //Hibernate don´t know that the user contains the food when persist the
  // food, because the relationship is unidirectional, therefore:
  sess = HibernateUtil.getSessionFactory().openSession();
  tx = sess.beginTransaction();
  sess.update(user);
  tx.commit();
  sess.close();  

  //Then I retrieve the user from DB
  sess = HibernateUtil.getSessionFactory().openSession();
  User userRetrieved =(User) sess.createCriteria(User.class)
             .add(Restrictions.idEq("@someone")).uniqueResult();
  sess.close();


  //Finally assert
  assertEquals(user.getFoods(), userRetrieved.getFoods());

}