@OneToOne或@ManyToOne在x.person上引用一个未知的实体：y.Person - 继承问题 [英] @OneToOne or @ManyToOne on x.person references an unknown entity: y.Person - inheritance issue

问题描述

我的Hibernate架构出了问题：

我有一个MappedSuperClass Person，一个Employee和一个Customer。

   - > Person.class 
 @MappedSuperclass 
 @Audited 
 public class Person extends PersistentObject {
 
 @Column（name =TITLE）
 @Enumerated（EnumType .STRING）
私人标题标题; 
 
 @Column（name =FIRST_NAME）
 private String fname = null; 
 
 @Column（name =LAST_NAME）
 private String lname = null; 
 
  - > Employee.class 
 @Entity 
 @Table（name =TBL_EMPLOYEE）
 @Audited 
 public class Employee extends Person {
 
  - > Customer.class 
 @Entity 
 @Table（name =TBL_CUSTOMER）
 @Audited 
 public class Customer extends Person {
  

这很好，但现在我的问题是：我已经使用项目列表（例如地址/联系人信息/ etc）扩展'Person'，并使用所有地址相同的单一表（所有地址）。

仅将地址添加到客户时，这是没有问题的：

   - > Customer.java 
 @OneToMany（cascade = {CascadeType.ALL}，mappedBy =customer）
 @LazyCollection（LazyCollectionOption.FALSE）
 private List< Address>地址= null; 
 
  - > Address.java 
 @Entity 
 @Table（name =TBL_ADDRESSES）
 @Audited 
 public class Address extends PersistentObject {
 @ManyToOne（fetch = FetchType.LAZY ）
 @JoinColumn（name =CUSTOMER_ID）
私人客户客户; 
  

现在我想将地址扩展到Employee。我认为，我可以使用超类型来保存地址，但这似乎并不正确：

   - > Person.java 
 @OneToMany（cascade = {CascadeType.ALL}，mappedBy =person）
 @LazyCollection（LazyCollectionOption.FALSE）
 private List< Address>地址= null; 
 
  - > Address.java 
 @Entity 
 @Table（name =TBL_ADDRESSES）
 @Audited 
 public class Address extends PersistentObject {
 @ManyToOne（fetch = FetchType.LAZY ）
 @JoinColumn（name =PERSON_ID）
私人人物; 
  

运行时，出现以下错误： @OneToOne或@ManyToOne on at.test.Address.person引用一个未知实体：at.test.Person


是否有可能使两个实体使用相同的表的地址？ ID不会冲突，因为员工和客户共享相同的顺序。或者我应该以另一种方式设计这个问题？

非常感谢您的提前和最好的问候！

解决方案

从Flo的链接中，我意识到用于解决问题的正确范例是每个子类一个表。感谢您的帮助！

i've got a problem with my Hibernate Architecture:

I have a MappedSuperClass Person, and an Employee and a Customer.

--> Person.class
@MappedSuperclass
@Audited
public class Person extends PersistentObject {

    @Column(name="TITLE")
    @Enumerated(EnumType.STRING)
    private Title title;

    @Column(name="FIRST_NAME")
    private String  fname       = null;

    @Column(name="LAST_NAME")
    private String  lname       = null;

--> Employee.class
@Entity
@Table(name="TBL_EMPLOYEE")
@Audited
public class Employee extends Person {

--> Customer.class
@Entity
@Table(name="TBL_CUSTOMER")
@Audited
public class Customer extends Person {

This works well, but now to my Problem: I've extended 'Person' with an List of Items, eg Addresses/Contact Informations/etc, and use a Single Table for all Items of the same Type (all Addresses).

When adding the Address to Customer only, this is no problem:

--> Customer.java
@OneToMany(cascade = {CascadeType.ALL}, mappedBy="customer")
@LazyCollection(LazyCollectionOption.FALSE)
private List<Address> addresses = null;

--> Address.java
@Entity
@Table(name="TBL_ADDRESSES")
@Audited
public class Address extends PersistentObject {
    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="CUSTOMER_ID")
    private Customer customer;

Now i'd like to extend the addresses to the Employee also. I thought, that i could use the Supertype to hold the address, but this doesn't seem to work right:

--> Person.java
@OneToMany(cascade = {CascadeType.ALL}, mappedBy="person")
@LazyCollection(LazyCollectionOption.FALSE)
private List<Address> addresses = null;

--> Address.java
@Entity
@Table(name="TBL_ADDRESSES")
@Audited
public class Address extends PersistentObject {
    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="PERSON_ID")
    private Person person;

When running, i get the following error: @OneToOne or @ManyToOne on at.test.Address.person references an unknown entity: at.test.Person

Is there a possibility to have both entities use the same table for their addresses? IDs wont conflict, because Employee and Customer share the same sequence. Or should i design this problem in another way?

Thank you very much in advance and best regards!

解决方案

From Flo's link I realized that the correct paradigm to use for my problem was the "one table per subclass" one. Thanks for the help!

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