Hibernate JPA：mappedBy 引用异常

Question

我想用 Hibernate JPA 映射以下类：

我的代码如下所示：

   @Entity
    public class Grid{
        @Id
        @GeneratedValue
        private Long id;
        @Column(unique=true)
        private String name;
        private String location;
        private BigDecimal costsPerCPUMinute;

        @OneToMany(mappedBy="grid")
        private List<Membership> mem;

            public List<Membership> getMem() {
             return mem;
            }
            public void setMem(List<Membership> mem) {
            this.mem = mem;
            }

@Entity
public class User extends Person{
    @Column(nullable=false, unique=true)
    private String username;
    @Column(length=16,columnDefinition="BINARY(16)") 
    private byte[] password;

    @OneToMany(mappedBy="user")
    private List<Membership> mem;

    public List<Membership> getMem() {
        return mem;
    }
    public void setMem(List<Membership> mem) {
        this.mem = mem;
    }


@SuppressWarnings("serial")
@Entity
public class Membership  implements Serializable{

    private Date registration;
    private Double discount;    
    @Id
    @ManyToOne
    private Grid grid;
    @Id
    @ManyToOne
    private User user;

    public Grid getGrid() {
        return grid;
    }
    public void setGrid(Grid grid) {
        this.grid = grid;
    }
    public User getUser() {
        return user;
    }
    public void setUser(User user) {
        this.user = user;
    }

不幸的是，我收到以下异常：

Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown target entity property: dst1.model.Membership.grid in dst1.model.Grid.mem

据我了解消息，在会员资格中找不到网格。但是正如你在代码中看到的那样，在 Membership 类中肯定有一个名为 grid 的变量。

有人知道出了什么问题吗？

更新：根据 cmets 中的建议，我还尝试使用 IDClass 或 EmbeddedID 更改成员资格类。 EmbeddedID 版本如下所示：

 @SuppressWarnings("serial")
    @Entity    
public class Membership  implements Serializable{

        private Date registration;
        private Double discount;    
        @EmbeddedId
        private MembershipPK membershipPK;

        public Membership(){};

        public MembershipPK getMembershipPK() {
            return membershipPK;
        }

        public void setMembershipPK(MembershipPK membershipPK) {
            this.membershipPK = membershipPK;
        }

    @SuppressWarnings("serial")
    @Embeddable
    public class MembershipPK implements Serializable{
        @ManyToOne
        private Grid grid;
        @ManyToOne
        private User user;


        public Grid getGrid() {
            return grid;
        }
        public void setGrid(Grid grid) {
            this.grid = grid;
        }
        public User getUser() {
            return user;
        }
        public void setUser(User user) {
            this.user = user;
        }
    }

不幸的是，我仍然遇到同样的异常。

更新 2：我将在明天晚上从头开始重写所有三个课程，如果有任何改变，我会更新这篇文章。

Answer 1

你应该可以使用类似的东西

@Embeddable
public class MembershipId
{
       protected Grid grid;
       protected User user;
}

@Entity
public class Membership {
    @EmbeddedId
    MembershipId id;
}

@Entity
public class User {
    @OneToMany(mappedBy="id.user")
    private Set<Membership> memberships = new HashSet<Membership>();
}

Answer 2

在我的脑海中：这不应该是相当的

public class MembershipId implements Serializable {
  private Grid grid;
  private User user;

  // immutable constructor, getters, equals, and hashCode
}

@Entity
@IdClass(MembershipId.class)
public class Membership implements Serializable {

  @Id @ManyToOne
  private Grid grid;

  @Id @ManyToOne
  private User user;

  // rest of class
}

编辑：异常告诉您的是，您的 Grid 类有一个名为 mem 的字段，并且该字段表示的实体需要一个网格字段，但没有。这是您的 Grid.mem 需要网格字段的地方：

@Entity
public class Grid{
...
  @OneToMany(mappedBy="grid")
  private List<Membership> mem;

只有在 Membership.class 中有属性网格时，此映射才有效。如果将网格隐藏在 IdClass 中，Grid.mem 将找不到它。你可以试试这个：

@Embeddable
public class MembershipId implements Serializable {
  private Grid grid;
  private User user;

  // immutable constructor, getters, equals, and hashCode
}

@Entity
public class Membership implements Serializable {

  @EmbeddedId
  private MembershipId id;

  @ManyToOne
  private Grid grid;

  @ManyToOne
  private User user;

  // rest of class
}