server side validation in the main view cause also validation in partial view, how to stop that? - asp.net-mvc-3

Question

I have layout with login partial view (username, password and submit button) and models ( some controls and submit button)with validation(server side and client side) displayed in normal views (#RenderBody() in layout).
My problem is when do server side validation in any of my views it also validate the login partial view because it execute the httppost function of the login. how can I stop that??
login view controller
[HttpGet]
public ActionResult LogOn()
{
return PartialView();
}
//
// POST: /Account/LogOn
[HttpPut]
public ActionResult LogOn(LogOnModel model)
{
if (ModelState.IsValid)
{
if (MembershipService.ValidateUser(model.UserName, model.Password))
{
FormsService.SignIn(model.UserName, model.RememberMe);
ViewBag.UserName = model.UserName;
}
else
{
ModelState.AddModelError("", Resources.Account.Account.LoginFailureText);
}
}
return PartialView(model);
}
and model controller
public ActionResult MyModel()
{
ViewBag.DisplayThxMsg = false;
return View();
}
[HttpPost]
public ActionResult MyModel(Models.FeedbacksModel feedback)
{
if (ModelState.IsValid)
{
//do something
}
else{
//do another thing
}
return View(feedback);
}

Answer 1

I find your question very difficult to understand. Im guessing your problem is you have a login partial control displayed as part of site layout and is shown on all pages. So while submitting any page, the username password validation kicks in, and you want to prevent that.
Understand that all validation # server - side happens while model binding, As properties are bound to the posted fields, the attributes on the fields are looked at and honored / catered to. So to prevent server side validation simply put the login partial view in it's own form so it is not sent while submitting other forms on the page.
In short have 2 forms - one form for login and one for feedback. Don't put all input fields in the same form.
If you still have validation errors after that, then it is because of other reasons like, type conversion problems. The default model binder will add some errors for basic type conversion issues (for example, passing a non-number for something which is an "int").The sample DataAnnotations model binder will fill model state with validation errors taken from the DataAnnotations attributes on your model.
EDIT
If you look at line number 125
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
You have the above code which will render the login form.
It will do so inside the other form at line 45
<form id="form1" runat="server" method="post">
This has no end tag therefore it will encompass the whole document till </html>
You should change the structure from
<form id="form1" runat="server" method="post">
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
</form
to
<form id="form1" runat="server" method="post">
</form>
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
This line #using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");} will render this form <form id="LoginView1" action="LogOn"> and all child elements of it.
LATEST EDIT
In your layout page use this :
#Html.Partial("~/Views/Shared/LogOnPartial.cshtml", new LogOnModel())
instead of this :
#Html.Action("LogOnPartial", "Account")
The reason why it all works is, the LogOnPartial method marked with [HttpPost] is called because the request was in a POST context. What you want is, You just need the view without the action executing even when POSTing. The above code does that. It renders the view without calling the action method. MVC3 is sort of a stupid servent : It only knows that it should call the Action method marked with [HttpPost] when the request is in a post context. It doesn't know that the request is in a post context for another action (index) and not this one (logonpartial). So now you can remove this method
public ActionResult LogOnPartial()
{
return PartialView();
}
which will no longer be used.
Note that you need to change the account controller's LogOnPartial Method to return
return RedirectToAction("Index","Home"); instead of return PartialView(model); on successful login. And on FAILURE you cannot render a partialview as you have coded. You must return an entirely new View. It must neither be index nor LogonPartails - just return return View("Login_Error_View"); which has its own layout. Otherwise it will be difficult to control the workflow.