这是我的xml,需要将它转换为java。我用了JAXBjaxb unmarshalling with namespace
<?xml version="1.0"?>
<lm:order Id="PLG24M240U" JD="" aCount="2" SUCount="1" xmlns:lm="http://www.ae.com/Event/Load">
<lm:master>
<lm:ID>3</lm:ID>
<lm:Number>313</lm:Number>
<lm:ANumber>323</lm:ANumber>
</lm:master>
<lm:detail>
<lm:ID>3</lm:ID>
<lm:Number>3131</lm:Number>
<lm:ANumber>3232</lm:ANumber>
</lm:detail>
<lm:detail>
<lm:ID>3</lm:ID>
<lm:Number>3131</lm:Number>
<lm:ANumber>3232</lm:ANumber>
</lm:detail>
<lm:detail>
<lm:ID>3</lm:ID>
<lm:Number>313</lm:Number>
<lm:ANumber>323</lm:ANumber>
</lm:detail>
</lm:order>
和投掷以下异常 javax.xml.bind.UnmarshalException:意外的元素(URI: “http://www.ae.com/Event/Load”,地方: “订单”)。预计元素< {} LM:订单>
这是我的解组代码
jaxbContext = JAXBContext.newInstance(Order.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
Order order = (Order) jaxbUnmarshaller.unmarshal(file);
System.out.println(order);
订购POJO类
import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "lm:Order")
public class OrderPay {
@XmlAttribute
private String Id;
@XmlAttribute
private String JD;
@XmlAttribute
private String aCount;
@XmlAttribute
private String pCount;
/*@XmlElement
private Master master;
@XmlElement
private List<Detail> details = new ArrayList<Detail>();*/
}
能否请你帮我在读也,目前通过文件读取,需要阅读为XML字符串。
@isim:雅它是一个错字,在代码中没有编译时错误。谢谢 – Rosh 2014-09-13 06:10:32