写了一个通过数据库生成XML 程序 有点问题 估计不大

Question

用Iterator去迭代List呀那样比较安全！

Answer 1

public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) {
// TODO Auto-generated method stub
        List list=new ArrayList();
        Session session=HibernateSessionFactory.getSession();
Transaction tr=session.beginTransaction();
tr.commit();
 
String statement = "from User";
list = session.createQuery(statement).list();
    
    Element user;
    Element username;
    Element password;
    Iterator iterator=list.iterator();
    DocumentBuilderFactory documentbuilderfactory = DocumentBuilderFactory.newInstance();
    try{        
       DocumentBuilder docBuilder = documentbuilderfactory.newDocumentBuilder();
       Document doc = docBuilder.newDocument();
       Element rootElement = doc.createElement("USER");
       //while(iterator.hasNext()){
      user = doc.createElement("user");
      user.setAttribute("userid",iterator.next().toString());
      System.out.println(iterator.next().toString());
         rootElement.appendChild(user);          username = doc.createElement("username");
         username.appendChild(doc.createTextNode(iterator.next().toString()));
         user.appendChild(username);          password = doc.createElement("password");
         password.appendChild(doc.createTextNode(iterator.next().toString()));
         user.appendChild(password);         
       //}
       doc.appendChild(rootElement);
       //格式化XML 文档
       OutputFormat outformat = new OutputFormat(doc);
       //通过输出流创建XML
       OutputStream os = new FileOutputStream("c:\\cdn\\xml\\DbToXml.xml");
       // 通过XMLSerializer 生成XML 文档
       XMLSerializer xmlserilizer = new XMLSerializer(os,outformat);
       // 对文件进行写操作.
       xmlserilizer.serialize(doc);     }catch(Exception ex){
        ex.printStackTrace();
    }
  
 
 
return mapping.findForward("success");
}改了上面了
能生成XML但结果是：<?xml version="1.0" encoding="UTF-8"?>
<USER><user userid="com.sh.hibernate.User@14093f"><username>com.sh.hibernate.User@3a3b6b</username><password>com.sh.hibernate.User@112fc47</password></user></USER>