【发布时间】:2014-11-07 07:41:11
【问题描述】:
这是我的xml,需要转换成java。我用过jaxb
<?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>
并抛出以下异常 javax.xml.bind.UnmarshalException:意外元素(uri:“http://www.ae.com/Event/Load”,本地:“订单”)。预期元素是
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>();*/
}
能否请您帮助我阅读,目前正在阅读文件,需要读取为 XML 字符串。
【问题讨论】:
-
@isim :是的,这是一个错字,代码中没有编译时错误。谢谢
标签: java xml-parsing namespaces jaxb