MVC Bootstrap modal show validation result - asp.net-mvc

Question

The problem:
I want to preform submit to form that is in modal and if validation faild to get the error message on the modal.
I'm using ajax validation (jQuery) as detailed here
Is there an elegant way to perform submit but on faild stay at modal to show error message?
My code:
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Create(Cam c)
{
ViewBag.id = c.id;
using (Entities db = new Entities())
{
if (ModelState.IsValid)
{
db.camp.Add(c);
db.SaveChanges();
return RedirectToAction("Index", new { id = c.id });
}
}
return null;
}
Client:
#using (Html.BeginForm("Create", "Camp", FormMethod.Post, new { model =
Model }))
{
#Html.AntiForgeryToken()
<dt>
name:
</dt>
<dd>
#Html.TextBoxFor(model => model.name, new { #class = "form-control", #placeholder = "name", #id = "txtVenueID", style = "width:150px" })
</dd>
<dd>
#Html.ValidationMessageFor(model => model.name)
</dd>
<div class="modal-footer ">
<button type="button" class="btn btn-default" data-
dismiss="modal">Close</button>
<button type="submit" class="btn btn-primary">Save</button>
</div>
}
Model:
public partial class Cam
{
[Display(Name = "Name")]
[Required(ErrorMessage = "Require {0}")]
string name { get; set; }
}

Answer 1

Show the first form in modal via the action that build it but keep the model state errors - use This
For view to submit the form but keep result on the modal use the following:
<button type="submit" class="btn btn-primary">Save</button>
Javascript to get the submit result into the modal:
$(function () {
$.ajaxSetup({ cache: false });
$(':submit[data-modal]').on("click", function (e) {
e.preventDefault();
var linkObj = $(this).closest('form');
$.ajax({ // create an AJAX call...
data: linkObj.serialize(), // get the form data
type: linkObj.attr('method'), // GET or POST
url: linkObj.attr('action'), // the file to call
success: function(response) { // on success..
$('#Modal-Content').html(response); // update the DIV
}
});
});
});