忽略JAXB中的根元素

Question

我想使用JAXB解析xml文件。 我的问题是，我需要跳过根节点，如果我从xml文件中删除它，我得到我所需要的，否则 - 我得到一个空的对象。

我给一个简单的XML和我的代码（它的工作方式）：

XML：

<?xml version="1.0" encoding="utf-8"?> 
<Root> 
<!-- <Root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:noNamespaceSchemaLocation="Office.xsd"> 
--> 
    <Office> 
     <Employees> 
      <Employee> 
       <name>George</name> 
       <rank>3</rank> 
      </Employee> 
      <Employee> 
       <name>Michael</name> 
       <rank>5</rank> 
      </Employee> 
      <Employee> 
       <name>Jeff</name> 
       <rank>1</rank> 
      </Employee> 
      <Employee> 
       <name>Todd</name> 
       <rank>7</rank> 
      </Employee> 
      <Employee> 
       <name>Jessica</name> 
       <rank>5</rank> 
      </Employee> 
     </Employees> 
    </Office> 
</Root> 

办公类：

import javax.xml.bind.annotation.*; 
import java.util.Vector; 

@XmlRootElement 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Office { 

    @XmlElementWrapper(name = "Employees") 
    @XmlElement(name = "Employee") 
    private Vector<Employee> employees; 

} 

Employee类：

import javax.xml.bind.annotation.XmlElement; 
import javax.xml.bind.annotation.XmlAccessorType; 
import javax.xml.bind.annotation.XmlAccessType; 

@XmlAccessorType(XmlAccessType.FIELD) 
public class Employee { 

    @XmlElement(name="name") 
    private String name; 
    @XmlElement(name="rank") 
    private int rank; 


    public void promote() { 
     rank++; 
    } 

} 

Driv呃：

import javax.xml.stream.*; 
import javax.xml.bind.JAXBContext; 
import javax.xml.bind.JAXBException; 
import javax.xml.bind.Unmarshaller; 
import javax.xml.bind.Marshaller; 

import java.io.FileReader; 
public class Driver { 

    public static void main (String[] args) { 
     parseOffice(); 

    } 

    public static void parseOffice() { 
     try { 

      XMLInputFactory f = XMLInputFactory.newInstance(); 
      XMLStreamReader reader = f.createXMLStreamReader(new FileReader("Office.xml")); 

      JAXBContext context = JAXBContext.newInstance(Office.class); 
      Unmarshaller unmarshaller = context.createUnmarshaller(); 
      Office office = unmarshaller.unmarshal(reader, Office.class).getValue(); 

      Marshaller marshaller = context.createMarshaller(); 
      marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true); 
      marshaller.marshal(office, System.out); 

     } 
     catch (Exception e) { 
      e.printStackTrace(); 
     } 
     } 
} 

Answer 1

您可以用StAX的XMLStreamReader解析XML，然后推进它到你想解组的元素，然后解组它。

我贴一个完整的例子，应有助于对相关问题链接如下：

• How to unmarshall SOAP response using JAXB if namespace declaration is on SOAP envelope?

Answer 2

你为什么不创建一个通用的根元素？

@XmlRootElement(name="Root" ...) 
public class Root { 
    @XmlAnyElement(lax=true) 
    private Object content; 
} 

将它添加到您的上下文和解组。你应该得到一个JAXBElement<Office>作为content。

Answer 3

只需添加层级的根类即可。并从Root类获取Office类。

根类： -

import javax.xml.bind.annotation.*; 
import java.util.Vector; 

@XmlRootElement 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Root { 


@XmlElement(name = "Office") 
private Office office; 
} 

办公室类

import javax.xml.bind.annotation.*; 
import java.util.Vector; 


    @XmlAccessorType(XmlAccessType.FIELD) 
    public class Office { 

    @XmlElementWrapper(name = "Employees") 
    @XmlElement(name = "Employee") 
    private Vector<Employee> employees; 

    } 

变化解析方法： -

 JAXBContext context = JAXBContext.newInstance(Root.class); 
     Unmarshaller unmarshaller = context.createUnmarshaller(); 
     Root root = unmarshaller.unmarshal(reader, Root.class).getValue(); 
     Office office = root.getOffice();