忽略JAXB中的根元素

Question

我想使用JAXB解析xml文件。 我的問題是，我需要跳過根節點，如果我從xml文件中刪除它，我得到我所需要的，否則 - 我得到一個空的對象。

我給一個簡單的XML和我的代碼（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> 

辦公類：

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類：

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，然後推進它到你想解組的元素，然後解組它。

我貼一個完整的例子，應有助於對相關問題鏈接如下：

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

Answer 2

你爲什麼不創建一個通用的根元素？

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

將它添加到您的上下文和解組。你應該得到一個JAXBElement<Office>作爲content。

Answer 3

只需添加層級的根類即可。並從Root類獲取Office類。

根類： -

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

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


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

辦公室類

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; 

    } 

變化解析方法： -

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