在spring data jpa中使用@PrimaryKeyJoinColumn注解

Question

我尝试使用@PrimaryKeyJoinColumn 注释。执行此操作时，我收到一个错误 - 尝试从空的一对一属性分配 id。

用户已保存，但地址未保存。我想为User 和Address 表创建一个公共主键。我找到了一个例子here。

请告诉我我做错了什么，为什么这个例子对我不起作用？

https://github.com/mytestPercon/TestHiber

User.java

@Entity
@Table(name = "user", schema = "TestKeyJoin")
public class User implements Serializable {

   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   @Column(name = "id")
   private Long id;

   @Basic
   @Column(name = "name")
   private String name;

   @OneToOne(mappedBy = "user")
   @PrimaryKeyJoinColumn
   private Address activated;

   // Getter and Setter ...
}

Address.java

@Entity
@Table(name = "Address", schema = "TestKeyJoin")
public class Address implements Serializable {

   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   @Column(name = "id")
   private long id;

   @Basic
   @Column(name = "city")
   private String city;

   @OneToOne
   @MapsId
   @JoinColumn(name = "id")
   private User user;

   // Getter and Setter ...

}

SaveController.java

@Controller
public class SaveController {

   @Autowired
   ServiceJpa serviceJpa;

   @GetMapping(value = "/saveUser")
   public String getJpa () {
      User user = new User();
      user.setId(1L);
      user.setName("Michael Joseph Jackson");
      serviceJpa.saveUser(user);

      Address address = new Address();
      address.setId(1L);
      address.setCity("Los Angeles");
      serviceJpa.saveActivated(address);

      return "/saveUser";
   }
}

Answer 1

尝试通过以下方式更正您的映射：

@Entity
@Table(name = "user", schema = "TestKeyJoin")
public class User implements Serializable {

   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   @Column(name = "id")
   private Long id;

   @Column(name = "name")
   private String name;

   @OneToOne(mappedBy = "user", cascade = CascadeType.ALL)
   private Address address;

   // ...
}

@Entity
@Table(name = "Address", schema = "TestKeyJoin")
public class Address implements Serializable {

   @Id
   private Long id;

   @Column(name = "city")
   private String city;

   /* You can use @PrimaryKeyJoinColumn instead of the @MapsId here.
      See the Example 153. Derived identifier @PrimaryKeyJoinColumn
      from the hibernate documentation.
   */
   @OneToOne
   @MapsId
   @JoinColumn(name = "id")
   private User user;

   // ...
}

然后通过以下方式保存用户：

@GetMapping(value = "/saveUser")
public String getJpa () {

   User user = new User();
   user.setName("Michael Joseph Jackson");

   Address address = new Address();
   address.setCity("Los Angeles");

   // make both sides of the bidirectional @OneToOne in-sync
   user.setAddress(address);
   address.setUser(user);

   serviceJpa.saveUser(user);

   return "/saveUser";
}

还有几个注意事项：

1. 每当形成双向关联时，应用程序开发人员必须确保双方始终保持同步。

2. 你不应该单独保存你的实体，你应该在你的关联中使用正确的cascading。

3. 您不应为generated identifier 设置值。

4. 当您在Address.user 上使用@MapsId 时，实际上意味着Address 实体将从一对一关联中借用标识符。因此，您不应将@GeneratedValue 注释用于Address.id。 （见this）

Answer 2

删除地址实体中id字段上的 @GeneratedValue(strategy = GenerationType.IDENTITY)，因为它是用户实体的外键，不应自动递增。 编辑：尝试添加这些行

 public String getJpa () {

            User user = new User();
            user.setId(1L);
            user.setName("Michael Joseph Jackson");
            Address address = new Address();
            address.setId(1L);
            address.setCity("Los Angeles");
            user.setActivated(address);
            serviceJpa.saveUser(user);
             return "/saveUser";
        }