Making primary key same as foreign key in Hibernate entity

I am new to Hibernate. I am working on two entities as follows:

Entity 1 is as follows:

@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 ?

In the associated table in mysql i.e. vm_user, user_id is both primary key as well as foreign key which refers to id of user table associated with User entity.

Entity 2 is as follows:

@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;

In the associated table in mysql i.e. my_entity, primary key is a combination of id of a and id of b. I am not getting how to denote this in Hibernate entity MyEntity.

Regarding the same, I have gone through a few posts: https://stackoverflow.com/questions/7146671/hibernate-foreign-key-as-part-of-primary-key and https://stackoverflow.com/questions/58527109/jpa-hibernate-composite-primary-key-with-foreign-key, but no getting idea how to do these two ?

The solution is @MapsId

For example

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

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

    @MapsId
    @OneToOne
    private User user;  

You can also remove the @JoinColumn(unique = true) because @Id makes it unique already.

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;

Please find more information in the Hibernate documentation https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#identifiers-derived

You need to use @EmbeddedId and @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;  
}

You can do the same thing for MyEntity as below,

@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;
}

VM Users class

public class VmUser implements Serializable {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;
@OneToOne
@JoinColumn(name="ID")
private Users user;

Users class

public class Users implements Serializable {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;
@OneToOne(mappedBy="user")
private VmUser vmUser;

A class

@Entity
public class A implements Serializable {
@Id
private long id;
@OneToMany(mappedBy="a")
private List<MyEntity> myEntitys;

B Class

@Entity
public class B implements Serializable {
@Id
private long id;
@OneToMany(mappedBy="b")
private List<MyEntity> myEntitys;

MyEntity class

@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 class

@Embeddable
public class MyEntityPK implements Serializable {
@Column(insertable=false, updatable=false)
private long id1;
@Column(insertable=false, updatable=false)
private long id2;