使用同样的HTTPService如何获取更新的数据我有两个httpservice.one这从servlet访问数据和一个存储数据到另一个servlet柔性形式。 首先,当IM从正在工作的servlet访问数据和存储部分也working..so当我再次拨打访问servlet IM没有得到更新display..the访问servlet是没有得到再次调用.. 这是我的访问servlet代码在柔性
public void doPost(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException
{
PrintWriter out=response.getWriter();
try
{
response.setContentType("text/html");
String gradeName=request.getParameter("tx1");
System.out.println(gradeName);
gradeName=gradeName.toUpperCase();
Session session = HibernateUtil.getSessionFactory().openSession();
Transaction tx = session.beginTransaction();
Grade g=new Grade(gradeName);
session.save(g);
tx.commit();
session.close();
//HibernateUtil.shutdown();
out.println("Added Successfully");
}
catch(ConstraintViolationException e)
{
out.println("Grade is already Present");
}
catch(Exception e)
{
e.printStackTrace();
}
}
}
这是我显示的servlet
保护无效的doGet(HttpServletRequest的请求,响应HttpServletResponse的)抛出的ServletException,IOException异常{
Session session = HibernateUtil.getSessionFactory().openSession();
Transaction tx=session.beginTransaction();
Query q=session.createQuery("from Grade");
List l=q.list();
Grade t;
PrintWriter out=response.getWriter();
response.setContentType("text/xml");
String str="<?xml version=\"1.0\" encoding=\"utf-8\"?><top>";
for(int i=0;i<l.size();i++)
{
t=(Grade)l.get(i);
str+="<inside><id>"+t.getGradeId()+"</id>";
str+="<name>"+t.getGradeName()+"</name></inside>";
}
str+="</top>";
out.println(str);
System.out.println("yattaa->"+str);
tx.commit();
session.close();
HibernateUtil.shutdown();
请不要使用标签来缩进代码。使用4或2个空格。这可以在任何像样的编辑器/ IDE中配置。 – BalusC 2010-01-28 13:38:55