2

我正在编写一个 Resteasy 服务器应用程序,并且无法让我的超类编组。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,我得到如下内容:

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

没有从 Person 类继承的 name 字段的输出。

我究竟做错了什么?

谢谢,拉尔夫

4

1 回答 1

0

我不确定您如何配置 JAXB 上下文或编组器,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);

}

给出:-

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<employee>
    <name>Ralph</name>
    <id>1</id>
</employee>
于 2010-03-10T18:46:16.267 回答