Java 配置模型类以使对象转换为Xml将生成外键的详细信息

我有两个班名雇主和雇员

雇主：Java

@XmlRootElement(name = "Employer")
@XmlAccessorType(XmlAccessType.FIELD)
@Entity
@Table(name = "employer")
public class Employer implements Serializable{
    @Id
    @GeneratedValue
    @Column(name = "id", length = 10)
    private int id;

    @XmlElement(name = "companyName")
    @Column(name = "name", length = 50)
    private String companyName;

    // Getter and setter methods
}

Java

@XmlRootElement(name = "Employee")
@XmlAccessorType(XmlAccessType.FIELD)
@Entity
@Table(name = "employee")
public class Employee implements Serializable{
    @Id 
    @GeneratedValue
    @Column(name = "id", length=10)
    private int id;

    @XmlElement(name = "name")
    @Column(name = "name", length=50)
    private String name;

    @XmlElement(name = "email")
    @Column(name = "email", length=50)
    private String email;

    // What should I put here If I want the details of Employer class also to be converted into xml data while Object to XML conversion using JAXB
    @ManyToOne(cascade={CascadeType.PERSIST},fetch = FetchType.LAZY)
    @JoinColumn(name="employer_id")
    private Employer employer;

    // Getter and setter methods
}

通常，我会获取数据列表，然后将其转换为Json，然后用javascript解析Json并使用3;来完成。我会根据自己的需要来做每一件事。在这种情况下，如果我想得到雇主的名字，我可以简单地将其命名为

row.employer.name

（下划线JS）来使用这个名字。但由于这里是原始xml数据，我不知道该怎么做。如果我能学习如何用xml生成嵌套对象数据，那么我就可以继续，否则我必须手动创建对象到xml