登陆的时候 我想用 session 保存一个用户
HttpSession session = request.getSession();
UserForm userForm = (UserForm)form;
if("".equals(session.getAttribute("userForm"))){
userForm = (UserForm)form;
session.setAttribute("userForm",userForm);
}else{
userForm = (UserForm)session.getAttribute("userForm");
}
这样写session.getAttribute("userForm")会出错
后面 我想 出来
HttpSession session = request.getSession();
session.setAttribute("user","no");
UserForm userForm = (UserForm)form;
if(session.getAttribute("user").equals(no)){
userForm = (UserForm)form;
session.setAttribute("userForm",userForm);
session.setAttribute("user","yes");
}if else(session.getAttribute("user").equals(yes)){
userForm = (UserForm)session.getAttribute("userForm");
}
但我 总觉得 这样写不太好
大家都是 怎么思考的 ?
HttpSession session = request.getSession();
UserForm userForm = (UserForm)form;
if("".equals(session.getAttribute("userForm"))){
userForm = (UserForm)form;
session.setAttribute("userForm",userForm);
}else{
userForm = (UserForm)session.getAttribute("userForm");
}
这样写session.getAttribute("userForm")会出错
后面 我想 出来
HttpSession session = request.getSession();
session.setAttribute("user","no");
UserForm userForm = (UserForm)form;
if(session.getAttribute("user").equals(no)){
userForm = (UserForm)form;
session.setAttribute("userForm",userForm);
session.setAttribute("user","yes");
}if else(session.getAttribute("user").equals(yes)){
userForm = (UserForm)session.getAttribute("userForm");
}
但我 总觉得 这样写不太好
大家都是 怎么思考的 ?
既然一个用户登录了,你已经session.setAttribute("userForm",userForm); 了
判断这个userFrom存不存在就可以了吧?
if(session.getAttribute("userForm") != null){
userForm = (UserForm)form;
session.setAttribute("userForm",userForm);
}else{
userForm = (UserForm)session.getAttribute("userForm");
}
这样就应该没问题了
现在没了 呵呵
我是从login.jsp 到 userlist.jsp这个 页面
public class UserAction extends Action {
@Resource
UserService userService; @Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response)
throws Exception {
HttpSession session = request.getSession();
UserForm userForm; if (session.getAttribute("userForm") == null) {
userForm = (UserForm) form;
session.setAttribute("userForm", userForm);
} else {
userForm = (UserForm) session.getAttribute("userForm");
}
String kind = userService.getUserKind(userForm.getUserId(), userForm
.getUserPassword());
if (!"".equals(kind) && kind != null) {
session.setAttribute("kind", kind);
int maxresult = 4;
int firstindex = (userForm.getPage() - 1) * maxresult;
PageView<tbUser> pageView = new PageView<tbUser>(maxresult,
userForm.getPage());
pageView.setQueryResult(userService.getScrollData(maxresult,
firstindex)); request.setAttribute("pageView", pageView);
for (tbUser u : pageView.getRecords())
request.setAttribute("a", u.getuserKind());
return mapping.findForward("list");
} else if ("".equals(kind) && kind == null) {
request.setAttribute("message", "用户密码错误");
return mapping.findForward("failure");
}
request.setAttribute("message", "请先登录!");
return mapping.findForward("failure");
}
}
这样 userForm 是被我 保存 下来 但是 我 UserForm 类里面本来一个 PAGE 属性
来控制 第几页 的
但是 这个 又需要 变动的这个 怎么办!
应该是我 设计 的 问题 我是 抄 网上的
大家 是 怎么 设计的 ?
UserForm userForm = (UserForm)form;
session.setAttribute("userForm",userForm);没有了
我水平 还不够 呵呵
HttpSession session = request.getSession();
UserForm userForm = (UserForm) form; if (session.getAttribute("userForm") == null) {
session.setAttribute("userForm", userForm);
} else {
int tpage = userForm.getPage();
userForm = (UserForm) session.getAttribute("userForm");
userForm.setPage(tpage);
}
我 这样 改 就 OK 了
lz的意思是说,怎么让jsp页面动态的翻页么?
如果是的话,就将page属性的值传到这个action里面 重新查询下就ok了