Question

我正在尝试使用JAXB将obj“Response”转换为XML。但我不想将我的对象的一个数据成员（此处为“responseData”）添加到XML。但我得到一个例外 - “IlleagalAnnotationException”。属性responseData存在但未在@ XmlType.propOrder中指定这是我的代码..

@XmlRootElement(name = "Root")
@XmlType(propOrder = {"payload", "result"})
public abstract class Response
{
    private String payload;
    private String result;

    @XmlElement(name = "Payload")
    public String getPayload() {
      return payload;
    }
    public void setPayload(String payload) {
        this.payload = payload;
    }
    @XmlElement(name = "Result")
    public String getResult() {
        return result;
    }
    public void setResult(String result) {
        this.result = result;
    }

    public abstract String getResponseData() ;
    public abstract void setResponseData(String responseData) ;
}

Answer 1

添加@XmlTransient以忽略它。

@XmlTransient
public abstract String getResponseData();

JAXB编组异常

1 个答案: