ErrorClassException:对象[id=null]不属于指定的子类

Question

我有一个像下面这样的模型：

@Entity(name = "request")
public class VisitRequest {
  @Id
  @GeneratedValue
  private Long id;

  @OneToMany(mappedBy = "visitRequest", cascade = CascadeType.ALL, orphanRemoval = true)
  @JsonManagedReference
  private List<Visitor> visitors;

  //default constructor, getters and setters
}

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "class")
public class Visitor {
  @Id
  @GeneratedValue
  private Long id;

  @ManyToOne
  @JsonBackReference
  private VisitRequest visitRequest;

  //default constructor, getters and setters

}

@Entity
public class ContactPerson extends Visitor {

  private PhoneNumber phoneNumber;

  //default constructor, getters and setters

}

但是当我尝试通过将其中一个访问者与一个联系人交换来更新一个visitRequest，并尝试在一个CRUD存储库visit request repository . save(visit request)上执行该方法时；我得到了这个异常:

Servlet。路径为[]的上下文中servlet[dispatcherServlet]的service（）引发了异常[Request processing failed；嵌套异常为org.springframework.orm.ObjectRetrievalFailureException:对象[id=null]不属于指定的子类[cern.ais.visions.core.domain.visitor.visitor]：给定对象的类与持久副本的类不匹配；嵌套的异常是org.hibernate。ErrorClassException:对象[id=null]不属于指定的子类[cern.ais.visions.core.domain.visitor.visitors]：给定对象的类与具有根本原因的持久副本的类不匹配

也许问题在于数据库中的contact_person和visitor表使用了相同的id？

我该如何解决这个问题？我一直在寻找解决方案，但没有一个对我有效。

Answer 1

@Entity(name = "request")
public class VisitRequest {
  @Id
  @GeneratedValue
  private Long id;

  @OneToMany(mappedBy = "visitRequest", cascade = CascadeType.ALL, orphanRemoval = true)
  @JsonManagedReference
  private List<Visitor> visitors;

  //default constructor, getters and setters
}

@Entity(name = "visitor")
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn(name = "visitor_type", discriminatorType = DiscriminatorType.STRING, length = 16)
@DiscriminatorValue(value = "visitor")
@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "class")
public class Visitor {
  @Id
  @GeneratedValue
  private Long id;

  @ManyToOne
  @JsonBackReference
  private VisitRequest visitRequest;

  //default constructor, getters and setters

}

@Entity(name = "contact_person")
@Table(name = "contact_person")
@DiscriminatorValue(value = "contact_person")
public class ContactPerson extends Visitor {

  private PhoneNumber phoneNumber;

  //default constructor, getters and setters

}

你可以像contactPersonDao.save一样坚持联系人（contactPerson），你不能坚持带有访问者链接的字段。我不是专家，但它对我有用。如果访问者不联系人，你可以将他作为访问者坚持下去

Answer 2

我怀疑您遇到这个问题的原因是您的注释< code>@Id和@GeneratedValue没有被继承。如果您在ContactPerson中定义属性id，您将拥有一个生成的Id，这将不再是一个问题。

尝试将ContactPerson类更改为：

@Entity
public class ContactPerson extends Visitor {

  @Id
  @GeneratedValue
  private Long id;

  private PhoneNumber phoneNumber;    
  //default constructor, getters and setters

}

Answer 3

您可能没有在 ContactPerson 中初始化 visitRequest 引用。请看一下下面的配置，它适用于Spring Boot JPA。我使用龙目岛生成了 Getter 和 Setter。

这里有一个在Spring Boot实现的工作示例https://github . com/conscious observer/SpringBootJpaInheritance

@Data
@Entity(name = "request")
class VisitRequest {
    @Id
    @GeneratedValue
    private Long id;

    @OneToMany(mappedBy = "visitRequest", cascade = CascadeType.ALL, orphanRemoval = true)
    @JsonManagedReference
    private List<Visitor> visitors = new ArrayList<>();
}

@NoArgsConstructor
@Data
@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "class")
class Visitor {
    @Id
    @GeneratedValue
    private Long id;

    @ManyToOne
    @JsonBackReference
    private VisitRequest visitRequest;

    public Visitor(VisitRequest visitRequest) {
        this.visitRequest = visitRequest;
    }
}

@NoArgsConstructor
@Data
@ToString(callSuper = true)
@EqualsAndHashCode(callSuper = true)
@Entity
class ContactPerson extends Visitor {
    private String phoneNumber;

    public ContactPerson(VisitRequest visitRequest, String phoneNumber) {
        super(visitRequest);
        this.phoneNumber = phoneNumber;
    }
}