Hi i want to create a Registration form in umbraco 7.5 version. I created Registration Document Type with template and Created RegistrationController and Registration Surface controller. i was able to show the typedview with model inheriting from Rendermodel.

In my registration view i have @Html.RenderArchetypePartials(Model.Content.ContentBlocks) and i have archtetype control to render a form.

When i submit the form i want to show message in view that email already exists. how can i pass the same model which i got from submiting the registration page so that it can show the message and same form again.

public ActionResult Register(RegistrationFormModel model)
    {
     var memberContext = memberService.GetByEmail(model.Username);
    if (memberContext != null)
        {
            ModelState.AddModelError("Username", "The username/email is already registered.");
            TempData["Error"] = "The entered email address is already registered.";
            return  PartialView("Registration",model);
        }
//    else do registration
}

i was able to get page back but model is new
My registration archetype control has this

<section class="block-simple">
<div class="container">
    <div class="block-content">
        <header>
            <h2>
                <sub>@(Model.GetValue<string>("title"))!</sub>
                @(Model.GetValue<string>("summary"))
            </h2>
        </header>
        @Html.RenderAction("ArchetypeForm/RegistrationCtrlForm", new  RegistrationFormModel())
    </div>
</div>

and my Registration Template has this

@inherits Umbraco.Web.Mvc.UmbracoTemplatePage<ContentModels.Registration> 
 @using ContentModels =  Umbraco.Web.PublishedContentModels;
 @using Archetype.Models;
 @using Archetype.Extensions;
 @{
  Layout = "Master.cshtml";
  }
  @Html.RenderArchetypePartials (Model.Content.ContentBlocks)

Can Any one have better idea to do or correction.