Question

嗨我试图检索我的会话值并将其用于_layout.cshtml，但每次我得到null异常(Object reference not set to an instance of an object.)这里是我的代码。我究竟做错了什么。 layout.cshtml无法加载〜/ Views / Shared / _LayoutHome.cshtml布局。 beacouse它没有获得任何会话价值。

@{
    if(HttpContext.Current.Session["LoggedUserType"].ToString() == "superadmin" &&  HttpContext.Current.Session["DEPTNM"].ToString() == "AlchemyAdmin")
    {
        Layout = "~/Views/Shared/_Layout.cshtml";
    }
    else
    {
        Layout = "~/Views/Shared/_LayoutHome.cshtml";

    }    
}

这是我的控制器代码，我可以使用我的会话值。

[HttpPost]
        public ActionResult Index(LoginModel model)
        {


            if (ModelState.IsValid)
            {
                if (DataAccess.LoginDAL.AdminIsValid(model.LOGINID, model.LOGINPW))
                {

                    var result = (from n in db.AslUsercoDbSet
                        where n.LOGINID == model.LOGINID &&
                              n.LOGINPW == model.LOGINPW
                        select new
                        {
                           companyid= n.COMPID, 
                           userid =  n.USERID, 
                           deptname= n.DEPTNM, 
                           usertype = n.OPTP
                        }
                        );
                    foreach (var n in result)
                    {
                        Session["loggedCompID"] = n.companyid;
                        Session["loggedUserID"] = n.userid;
                        Session["LoggedDepartment"] = n.deptname;
                        Session["LoggedUserType"] = n.usertype;
                    }

                    FormsAuthentication.SetAuthCookie(model.LOGINID, true);
                    return RedirectToAction("Index", "Home");
                }
                {
                    ModelState.AddModelError("", "Invalid Username or Password");
                }
            }
            return View();
        }

我在这里添加了我的dataaccess图层代码。如果存在，则验证用户。

 internal static bool AdminIsValid(string username, string password)
        {
            bool authenticated = false;

            string query = string.Format("SELECT * FROM [AslUsercoes] WHERE LOGINID = '{0}' AND LOGINPW = '{1}'", username, password);

            SqlCommand cmd = new SqlCommand(query, conn);
            conn.Open();

            SqlDataAdapter da = new SqlDataAdapter(cmd);
            DataTable ds = new DataTable();
            da.Fill(ds);
 SqlDataReader sdr = cmd.ExecuteReader();
            authenticated = sdr.HasRows;
            conn.Close();
            return (authenticated);
        }

Answer 1

这取决于您如何为会话分配值。

Session [“LoggedUserType”] = value（块不能有using（）语句。因为它会在这种情况下处理你的值对象。）

请参阅以下链接。

MVC 4 Persist Data throughout website

第二，

您需要用户全线（Html.ViewContext.HttpContext）

@{
    if(Html.ViewContext.HttpContext.Current.Session["LoggedUserType"].ToString() == "superadmin" &&  Html.ViewContext.HttpContext.Current.Session["DEPTNM"].ToString() == "AlchemyAdmin")
    {
        Layout = "~/Views/Shared/_Layout.cshtml";
    }
    else
    {
        Layout = "~/Views/Shared/_LayoutHome.cshtml";

    }    
}

检索会话变量并将其用于MVC 4（razor）中的viewstart

1 个答案: