如何在JAXB的幫助下處理嵌套的xml標記？

如何在JAXB的幫助下處理這樣的結構？有一個父根目錄和許多其他的內容，它們保存有關許多對象（例如狀態）的信息。如何在JAXB的幫助下處理嵌套的xml標記？

<?xml version='1.0' encoding='utf-8'?> 
    <root> 
     <states> 
      <state> 
       <name>Alabama</name> 
       <id>al</id> 
      </state> 
      <state> 
       <name>Alaska</name> 
       <id>ak</id> 
      </state> 
      ... 
     </states> 
     <airports> 
      <airport> 
       ... 
      </airport> 
      <airport> 
       ... 
      </airport> 
      ... 
     </airports> 
     ... 
    </root>

如果狀態，我創建的類國家：

@XmlRootElement(name = "root") 
    @XmlAccessorType(XmlAccessType.FIELD) 
    public class States { 
     @XmlElement(name="states") 
     private List<State> states; 

     public List<State> getBrokers(){ 
      return states; 
     } 

     public void setBrokers(List<State> states) { 
      this.states = states; 
     } 
    }

和類國家：

@XmlAccessorType(XmlAccessType.FIELD) 
public class State { 
    @XmlElement(name = "name") 
    private String name; 

    @XmlElement(name = "id") 
    private String id; 
    public State(){} 
    public State(String id, String name) { 
     this.id = id; 
     this.name = name; 
}

但當我致電國對象

public class JaxbParser { 
    public Object getObject(File file, Class c) throws JAXBException { 
     JAXBContext context = JAXBContext.newInstance(c); 
     Unmarshaller unmarshaller = context.createUnmarshaller(); 
     Object object = unmarshaller.unmarshal(file); 

     return object; 
    } 

    public void saveObject(File file, Object o) throws JAXBException { 
     JAXBContext context = JAXBContext.newInstance(o.getClass()); 
     Marshaller marshaller = context.createMarshaller(); 
     marshaller.marshal(o,file); 
    } 
} 

public static void main(String[] args) throws JAXBException { 
    JaxbParser parser = new JaxbParser(); 
    States brol = (States) parser.getObject(new File("C:\\Users\\...\\1.xml"), States.class); 

    List<State> brokerList = brol.getStates(); 
    System.out.println(brol); 
}

的stateList只包含一個id和id的狀態name等於null。我在做什麼錯？

來源

2017-05-07 Stanislav Studzinskyi

您將需要定義一個自定義的Xml適配器。 –

你在哪裏遍歷'List brokerList'？ – Vitaliy

public static void main(String args[]) throws 
JAXBException { 
try { 
    File file = new File("....xml"); 
    JAXBContext jaxbContext = 
    JAXBContext.newInstance(Classname.class); 
    Unmarshaller jaxbUnmarshaller = 
    jaxbContext.createUnmarshaller(); 
    Classname objectName = (Classname) 
    jaxbUnmarshaller.unmarshal(file); 

    List<Classname> list = objectName.getStates(); 
    int i = 1; 
    for (State ans : list){ 
     ..... 
    } 

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

來源

2017-05-07 18:21:23

謝謝你的回答，但我使用了一個相似的結構，我只是忘了將它添加到代碼中，現在你可以看到它。問題不在此，不幸的是。 –

如何在JAXB的幫助下處理嵌套的xml標記？

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