HTTP Post request from different controller actions and ModelState - asp.net-mvc

Question

I have a weird need in an ASP.NET MVC 3 application which blocks my current progress. Here is the case:
I have a little search engine for the products and I render this search engine on multiple pages. This SE makes a HTTP POST request to product controller's search action. It fine till here.
Let's assume that I am on home controller's index action (/home/index). I make a search and check if ModelState.IsValid. As a result, it is not valid. So, I should return this back with the entered model (so that user won't lose the values) and model state errors. But when I do that I ended up with different URL (/product/search) as expected.
If I do a redirect, I lose the ModelState and cannot display error messages.
I have different solutions so far and they all look dirty. Any idea?
Edit
Here is a little project which demonstrates this:
This is the ProductController:
public class ProductController : Controller {
[HttpPost]
public ActionResult Search(SearchModel searchModel) {
if (ModelState.IsValid) {
//Do some stuff...
return RedirectToAction("Index", "SearchResult");
}
return View(searchModel);
}
}
This is the SearchModel:
public class SearchModel {
[Required]
public string ProductCategory { get; set; }
[Required]
public string ProductName { get; set; }
}
This is the *_SearchPartial*:
#model MvcApplication20.SearchModel
#using (Html.BeginForm("search", "product"))
{
#Html.EditorForModel()
<input type="submit" value="Search" />
}
And finally this is the Home controller Index action view which renders the *_SearchPartial*:
#{
ViewBag.Title = "Home Page";
}
<h2>#ViewBag.Message</h2>
#Html.Partial("_SearchPartialView")
Here, when I submit the form and if the model state fails, how should I proceed at the Product controller Search action?

Answer 1

Here, when I submit the form and if the model state fails, how should
I proceed at the Product controller Search action?
Normally in this case you should render the _SearchPartialView but not as a partial but as a full view with layout so that the user can fix his errors. No need to stay at Home/Index in this case:
[HttpPost]
public ActionResult Search(SearchModel searchModel) {
if (ModelState.IsValid) {
//Do some stuff...
return RedirectToAction("Index", "SearchResult");
}
// since we are returning a view instead of a partial view,
// the _SearchPartialView template should be displayed with the layout
return View("_SearchPartialView", searchModel);
}
And if you wanted to stay on the same page upon error you could use an AJAX call to perform the search. So you would AJAXify this search form and then in the success callback test the result of the Search action and based on it decide whether to refresh the partial in order to show the error or redirect to the results action using window.location.href:
something along the lines of:
$(document).on('submit', '#searchForm', function() {
$.ajax({
url: this.action,
type: this.method,
data: $(this).serialize(),
success: function(result) {
if (result.redirectTo) {
// no validation errors we can redirect now:
window.location.href = result.redirectTo;
} else {
// there were validation errors, refresh the partial to show them
$('#searchContainer').html(result);
// if you want to enable client side validation
// with jquery unobtrusive validate for this search form
// don't forget to call the .parse method here
// since we are updating the DOM dynamically and we
// need to reattach client side validators to the new elements:
// $.validator.unobtrusive.parse(result);
}
}
});
return false;
});
This obviously assumes that you have now wrapped the partial call in a div with id="searchContainer" and that you provided an id="searchForm" when generating the search form:
<div id="searchContainer">
#Html.Partial("_SearchPartialView")
</div>
and now the search action:
[HttpPost]
public ActionResult Search(SearchModel searchModel) {
if (ModelState.IsValid) {
//Do some stuff...
return Json(new { redirectTo = Url.Action("Index", "SearchResult") });
}
return PartialView("_SearchPartialView", searchModel);
}

Answer 2

As far as I know the ModelState is lost when doing a RedirectToAction, the solution would be to save the modelstate in the TempData one example of this, that I'm using is this:
http://weblogs.asp.net/rashid/archive/2009/04/01/asp-net-mvc-best-practices-part-1.aspx#prg
This is also discussed in various posts for instance MVC Transfer Data Between Views