Anycleexcence
New Member
I have a view where I'm using a specific layout.cshtml file, other than the main shared layout page: "_LayoutExCr"This is fine for the Get part of the controller:\[code\] // // GET: /Exhibitor/Create public ActionResult Create() { return View("Create","_LayoutExCr"); }\[/code\]This works fine - and displays the Create view with the specific _LayoutExCr "master" page.However, in my POST for the Create method, if the wrong access code is entered, I want to return to the same view, using the _LayoutExCr "master" page - but VS2012 Express underlines in Red: \[code\] return View(exhibitor, "_LayoutExCr");\[/code\]\[quote\] The best overloaded method match for 'System.Web.Mvc.Controller.View(string, string)' has some invalid arguments\[/quote\]\[code\] // // POST: /Exhibitor/Create [HttpPost] public ActionResult Create(Exhibitor exhibitor) { if (ModelState.IsValid) { if (exhibitor.AccessCode == "myaccesscode") { db.Exhibitors.Add(exhibitor); db.SaveChanges(); return RedirectToAction("Thankyou"); } else { ModelState.AddModelError("", "The Access Code provided is incorrect."); return View(exhibitor, "_LayoutExCr"); } } return View(exhibitor, "_LayoutExCr"); }\[/code\]Can anyone let me know how to return the model to the view, using that same layout page please?Thank you,Mark