使用JAXB将属性封送为类似列表的嵌套根元素类型

我想使用JAXB API保存一个医院列表,每个医院都有自己的服务列表。为此,我实现了以下类

 public class Wrapper<T> {
     private List<T> items = new ArrayList<T>();
        @XmlAnyElement(lax=true)
        public List<T> getItems() {
            return items;
        }
    }
/*************************************/
 @XmlAccessorType(XmlAccessType.FIELD)
    @XmlRootElement(name="hopital")
    public class Hopital {
    private int id;
    private String nom,adresse,categorie="";
    @XmlElement(name="service")
    private List<Service> services=new ArrayList<>();
    private static List<Hopital> liste=new ArrayList<>();
    static File springDir;
    static JAXBContext context;
    static BufferedWriter writer = null;

    //constructeurs 
    //getters and setters 

    public static void addHopital(Hopital ... hops) throws Exception{
        File hfile = new File("hopitaux.xml");

        Wrapper<Hopital> hopitaux = new Wrapper<Hopital>();
        for (int i = 0; i < hops.length; i++) {
            hopitaux.getItems().add(hops[i]);
        }
        writer = new BufferedWriter(new FileWriter(hfile));
    context =JAXBContext.newInstance(Wrapper.class,Hopital.class,Service.class);
        JAXBElement<Wrapper> element=new JAXBElement<Wrapper>(new 
        QName("hopitaux"), Wrapper.class,hopitaux);
        Marshaller m = context.createMarshaller();
        m.setProperty(Marshaller.JAXB_ENCODING, "iso-8859-15");
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        m.marshal(element, writer); 
        writer.close();
    }
    }
/*************************************/
   @XmlAccessorType(XmlAccessType.FIELD)
    @XmlRootElement(name="service")
    public class Service {
    private int numero;
    private String libelle;

    //constructeurs 
    //getters and setters 

     }
/*************************************/

中的代码 Hopital.addHopital() 使用以下结构工作并提供xml输出:

<?xml version="1.0" encoding="iso-8859-15" standalone="yes"?>
<hopitaux>
    <hopital>
        <id>0</id>
        <nom>A</nom>
        <adresse>aaaaaaaa</adresse>
        <categorie></categorie>
        <service>
            <numero>1</numero>
            <libelle>car</libelle>
        </service>
        <service>
            <numero>1</numero>
            <libelle>chg</libelle>
        </service>
    </hopital>
    <hopital>
        .....
    </hopital>    
</hopitaux>

但是我想通过为这个服务列表创建一个根元素来获得以下结构,如下所示:

<hopitaux>
    <hopital>
        <id>0</id>
        <nom>A</nom>
        <adresse>aaaaaaaa</adresse>
        <categorie></categorie>
        <services>
           <service>
              <numero>1</numero>
              <libelle>car</libelle>
           </service>
           <service>
              <numero>1</numero>
              <libelle>chg</libelle>
           </service>
        </services>
    </hopital>
    <hopital>
        .....
    </hopital>    
</hopitaux>
</pre>