Question

我的目标是将XML文件解组为一个Java对象。我的XML看起来像：

window.location.href = "..."

我只从info1-info3收到信息。 BusinessOperationEvent包含XML文件中的所有XMLElements。解析器类：

public

BusinessOperationEvent是用于从XML收集信息的Java对象：

<root>
   <info1>
   <info2>
   <info3>
       <nested1>
           <nested2>
               <info4>
               <info5>
           <nested2>
       <nested1>
<root>

XML：

public class Parser {

public static void main(String[] args){
    try {
        ClassLoader classLoader = Parser.class.getClassLoader();
        File xml = new File(classLoader.getResource("XML.xml").getFile());

        JAXBContext jaxbContext = JAXBContext.newInstance(BusinessOperationEvent.class);
        Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
        BusinessOperationEvent event = (BusinessOperationEvent) unmarshaller.unmarshal(xml);
        System.out.println(event);
    } catch (JAXBException e) {
        throw new RuntimeException("Error unmarshalling XML");
    }
}
}

我在Java对象中只获得了与Method XML标签相关的信息，而不仅仅是nulls。

Answer 1

您应该构建类层次结构等消息和事件类

@XmlRootElement(name = "SrvPutBusinessOperationEventNf")
public class BusinessOperationEvent
{
    @XmlElement(name = "RqUID")
    private String rqUID;
    @XmlElement(name = "RqTm")
    private String rqTm;
    @XmlElement(name = "sPName")
    private String sPName;
    @XmlElement(name = "SystemId")
    private String systemId;
    @XmlElement(name = "Method")
    private String method;
    @XmlElement(name = "Message")
    Message message


}
public class Message
{
    @XmlElement(name = "Event")
    Event event;
}

等等

解组嵌套的XML标记

1 个答案: