在 Hibernate 实体中使主键与外键相同

Question

我是 Hibernate 的新手。我正在研究以下两个实体：

实体1如下：

@Entity
@Table(name = "vm_user")
public class VmUser implements Serializable {

    private static final long serialVersionUID = 1L;

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

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

    @Column(name = "created_date")
    private Instant createdDate;

    @Column(name = "last_modified_date")
    private Instant lastModifiedDate;

    @OneToOne
    @JoinColumn(unique = true)
    private User user;             <--- HOW WILL I DENOTE THIS PRIMARY KEY OF VMUSER ENTITY ?

在mysql的关联表中，即vm_user，user_id既是主键也是外键，指的是与User实体关联的user表的id。

实体2如下：

@Entity
@Table(name = "my_entity")
public class MyEntity implements Serializable {

    private static final long serialVersionUID = 1L;

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

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

    @Column(name = "created_date")
    private Instant createdDate;

    @Column(name = "last_modified_date")
    private Instant lastModifiedDate;

    @ManyToOne
    private A a;

    @OneToOne
    @JoinColumn(unique = true)
    private B b;

在mysql的关联表my_entity中，主键是id of a和id of b的组合。我不知道如何在 Hibernate 实体 MyEntity 中表示这一点。

关于同样的，我已经浏览了一些帖子：Hibernate foreign key as part of primary key 和 JPA & Hibernate - Composite primary key with foreign key，但不知道如何做到这两个？

Answer 1

解决办法是@MapsId

例如

@Entity
@Table(name = "vm_user")
public class VmUser implements Serializable {

    @Id
    @Column(name = "user_id")    
    private Integer id;

    @MapsId
    @OneToOne
    private User user;  

您也可以删除@JoinColumn(unique = true)，因为@Id 已经使其独一无二。

public class MyEntityPk implements Serializable {

   private Integer aId;
   private Integer bId;

   // IMPORTANT: Override equals() and hashCode()
}


@IdClass(MyEntityPk.class)
@Entity
@Table(name = "my_entity")
public class MyEntity implements Serializable {

    @Id
    private Integer aId;
    @Id
    private Integer bId;

    @MapsId("aId")
    @ManyToOne
    private A a;

    @MapsId("bId")
    @OneToOne
    private B b;

请在 Hibernate 文档https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#identifiers-derived中找到更多信息

Answer 2

你需要使用@EmbeddedId和@MapsId，

@Entity
@Table(name = "vm_user")
public class VmUser implements Serializable {
   @Id
    @Column(name = "user_id")    
    private Integer id;

    @MapsId("user_id")
    @OneToOne
    private User user;  
}

你可以对 MyEntity 做同样的事情，如下所示，

@Embeddable
class BKey {
  private int aId;
  private int bId;
}


@Entity
@Table(name = "my_entity")
public class MyEntity implements Serializable {

    @EmbeddedId
    private BKey primaryKey;

    @MapsId("aId")
    @ManyToOne
    private A a;

    @MapsId("bId")
    @OneToOne
    @JoinColumn(unique = true)
    private B b;
}

Answer 3

VM 用户类

public class VmUser implements Serializable {

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

@OneToOne
@JoinColumn(name="ID")
private Users user;

用户类

public class Users implements Serializable {

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

@OneToOne(mappedBy="user")
private VmUser vmUser;

一个类

@Entity
public class A implements Serializable {

@Id
private long id;

@OneToMany(mappedBy="a")
private List<MyEntity> myEntitys;

B级

@Entity
public class B implements Serializable {

@Id
private long id;

@OneToMany(mappedBy="b")
private List<MyEntity> myEntitys;

MyEntity 类

@Entity
public class MyEntity implements Serializable {
private static final long serialVersionUID = 1L;

@EmbeddedId
private MyEntityPK id;


@ManyToOne
@JoinColumn(name="ID1")
private A a;

@ManyToOne
@JoinColumn(name="ID2")
private B b;

MyEntityPK 类

@Embeddable
public class MyEntityPK implements Serializable {

@Column(insertable=false, updatable=false)
private long id1;

@Column(insertable=false, updatable=false)
private long id2;