JAXB编组超类

时间:2010-03-10 18:21:30

标签: jaxb marshalling jax-rs

我正在编写一个Resteasy服务器应用程序,但我很难让我的超级课程编组。我有这样的代码:

@XmlAccessorType(XmlAccessType.NONE)
@XmlRootElement(name = "person")
class Person {
  protected String name;

  @XmlElement(name = "name")
  public String getName() { return name; }

  public void setName(String name) { this.name = name; }
}

@XmlAccessorType(XmlAccessType.NONE)
@XmlRootElement(name = "employee")
class Employee extends Person {
  protected Integer id;

  @XmlElement(name = "id")
  public Integer getId() { return id; }

  public void setId(Integer id) { this.id = id; }
}

当我将Employee类编组为XML时,我得到这样的结果:

<employee>
  <id>12345</id>
</employee>

没有继承自Person类的名称字段的输出。

我做错了什么?

谢谢,拉尔夫

1 个答案:

答案 0 :(得分:0)

我不确定您是如何配置JAXB上下文或marshaller的,但以下内容: -

public static void main(String[] args) throws Exception
{

        Employee employee = new Employee();
        employee.setId(1);
        employee.setName("Ralph");

        JAXBContext context = JAXBContext.newInstance(Employee.class);
        Marshaller marshaller = context.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(employee, System.out);

}

得到: -

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<employee>
    <name>Ralph</name>
    <id>1</id>
</employee>