Question

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)
    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 {

这很有效，但现在我的问题是：我已经扩展了人物＆＃39;带有项目清单，例如地址/联系信息/等，并对同一类型的所有项目使用单表（所有地址）。

仅将地址添加到客户时，这不是问题：

--> 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;

现在我想将地址扩展到员工。我想，我可以使用Supertype来保存地址，但这似乎不能正常工作：

--> 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;

运行时，我收到以下错误：@OneToOne or @ManyToOne on at.test.Address.person references an unknown entity: at.test.Person

是否有可能让两个实体对他们的地址使用相同的表？ ID不会发生冲突，因为Employee和Customer共享相同的序列。或者我应该以另一种方式设计这个问题？

非常感谢您提前和最好的问候！

Answer 1

您必须将您的超类声明为@Entity而不是@MappedSuperClass和单表继承策略，因此您可以在ManyToOne关系中引用它

@Entity
@Table(name="PERSON")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="discriminator",discriminatorType=DiscriminatorType.STRING)
@DiscriminatorValue(value="P")
public abstract class Person {

    @Id
    @GeneratedValue
    @Column(name = "PERSON_ID")
    private Long personId;

    private String firstname;

    private String lastname;

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

@Entity
@Table(name="PERSON")
@DiscriminatorValue("E")
public class Employee extends Person {

}

@Entity
@Table(name="PERSON")
@DiscriminatorValue("C")
public class Customer extends Person {

}

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

Answer 2

从Flo的链接我意识到用于我的问题的正确范例是“每个子类一个表”一个。谢谢你的帮助！

Answer 3

在Address类中，您必须将参数insertable和updatable设置为false。它实际上是有道理的，因为如果你不这样做。 JPA允许您创建一些“Person”（可能未映射）对象，将其分配给地址，然后与地址对象一起保留。尝试使用以下代码：

@Entity
@Table(name="PERSON")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="discriminator",discriminatorType=DiscriminatorType.STRING)
public abstract class Person {

    @Id
    @GeneratedValue
    @Column(name = "PERSON_ID")
    private Long personId;

    private String firstname;

    private String lastname;

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

@Entity
@Table(name="PERSON")
@DiscriminatorValue("E")
public class Employee extends Person {

//some fields
}

@Entity
@Table(name="PERSON")
@DiscriminatorValue("C")
public class Customer extends Person {
    //some fields
}

@Entity
@Table(name="TBL_ADDRESSES")
@Audited
public class Address extends PersistentObject {
     @ManyToOne(fetch=FetchType.LAZY)
     @JoinColumn(name="PERSON_ID", insertable = false, updatable = false)
     private Person person;
}

@OneToOne或x.person上的@ManyToOne引用了一个未知实体：y.Person - 继承问题

3 个答案: