使用同樣的HTTPService如何獲取更新的數據我有兩個httpservice.one這從servlet訪問數據和一個存儲數據到另一個servlet柔性形式。 首先,當IM從正在工作的servlet訪問數據和存儲部分也working..so當我再次撥打訪問servlet IM沒有得到更新display..the訪問servlet是沒有得到再次調用.. 這是我的訪問servlet代碼在柔性
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();
}
}
}
這是我顯示的servlet
保護無效的doGet(HttpServletRequest的請求,響應HttpServletResponse的)拋出的ServletException,IOException異常{
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();
請不要使用標籤來縮進代碼。使用4或2個空格。這可以在任何像樣的編輯器/ IDE中配置。 – BalusC 2010-01-28 13:38:55