请高手帮忙想一下,一个JSP页面的问题(如果学生登陆就显示学生信息,老师登陆就显示老师的~~~谢谢,在线等

来源：百度知道 编辑：UC知道 时间：2024/06/08 07:47:44
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) {
GongyongForm gongyongForm = (GongyongForm) form; HttpSession session = request.getSession();
//从登陆页面请求一个登陆帐号过来
String id = (String)session.getAttribute("pwduser");
//学生管理
Usermanager usermanager = new Usermanager();
//老师管理
Teachermanager teachmanager = new Teachermanager();
//把输入帐号进入数据库进行叛断
TbUser user = usermanager.select(id);
TbTeacher tea = teachmanager.select(id);
//叛断结果
if(tea != null){
//如果叛断该人登陆的是老师,把信息set到老师类里面去
//跟上面一样,,不写了
gongyongForm.setGetfem(tea.getPinfen());
//我本来想到页面进行身份的叛断,在插入数据的时候我都设置了,如果是学生bodyfen="学生",老师的话就是bodyfen="老师"
session.setAttribute("bodyfen", tea.getBodyfen());
//跳到main.jsp页面显示信息
return mapping.findForward("ok");
}els

这个不需要在要页面上改...你都在登入是判断了!!!
那你在写一个Class:

public ActionForward 显示(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) {
List result = null;
HttpSession session = request.getSession();
String userName = (String) session.getAttribute("登入用户");
try {
Session ses = HibernateSessionFactory.getSession();

Criteria criteria = ses
.createCriteria("包内位置");
criteria.add(Expression.eq("actionBy", userName));

result = criteria.list();

HibernateSessionFactory.closeSession();
request.setAttribute("result", result);

} catch (HibernateException e) {
e.printStackTrace();
throw e;
}
return mapping.findForward("myTaskList");
}