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executing mvc controller using either Jquery or Ajax call

I am trying to mimic an mvc ActionLink. I want the whole row to be clickable. when the actionlink is clicked, it calls the connected controller and executes the code within. I want my Jquery/ajax call to do the same.
I've tried multiple ways of doing this with no luck. I'm currently at a point where the row is clickable and the Jquery sees that, however the ajax call does not execute. Or, if it does, the controller does not execute correctly
Here is the Jquery code that catches the click.
$(document).ready(function () {
$('#policyTable').on('click', '.clickable-row', function (event) {
$(this).addClass('primary').siblings().removeClass('primary');
var Id = $(this).closest('tr').children('td:first').text();
var url = "/Home/ReviewPolicy";
var uc = $(this).closest('tr').children('td:first').text();
alert("Does the click work? " + Id);
$.ajax({
type: "POST",
url: "/Home/ReviewPolicy",
dataType: 'text',
data: { Id: Id }
});
})
})
Here is the controller it is calling:
[HttpPost]
public ActionResult ReviewPolicy(string Id)
{
//Declare policyVM for individual policy
PolicyRenewalListVM model;
int val = Convert.ToInt32(Id);
using (Db db = new Db())
{
//Get the row
PolicyRenewalListDTO dto = db.RenewalPolicies.Find(val);
//confirm policy exists
if (dto == null)
{
return Content("This policy cannot be found.");
}
//initialize the PolicyRenewalListVM
model = new PolicyRenewalListVM(dto);
}
//return view with model
return View(model);
}
When the actionlink itself is clicked (It's not here in the code as I didn't see it being necessary for this problem) it fires and everything works as it should, but the jquery call, sending the very same value (Id) does not.

Still need to TryUpdateModel when posting through Ajax

just wondering if there is any point validating a view model using TryUpdateModel(), when calling a Post method through Ajax. I have the following code which is called via a jquery Ajax post:
[AcceptVerbs(HttpVerbs.Post)]
public JsonResult SubmitForm(ViewModel viewModel)
{
var valid = TryUpdateModel(viewModel);
if (valid)
{
var service = new Service();
var result = _tmpRepository.ExecuteService(viewModel));
return Json(new { Valid = valid, Response = result });
}
return Json(new { Valid = valid });
}
The way I see it, I'm using unobtrusive validation with data annotation on my view model. So the only way that validation wouldn't occur is if javascript was disabled and if javascript is disabled my Ajax request isn't going to do a fat lot!

You still need to validate any data server side. There's nothing stopping someone from submitting the request themselves, or modifying the parameters of the ajax call in the browser. Client side validation should only be a convenience for the user.
You can avoid the TryUpdateModel() though:
[AcceptVerbs(HttpVerbs.Post)]
public JsonResult SubmitForm(ViewModel viewModel)
{
if (ModelState.IsValid)
{
var service = new Service();
var result = _tmpRepository.ExecuteService(viewModel));
return Json(new { Valid = valid, Response = result });
}
return Json(new { Valid = valid });
}

jQuery to pass to MVC asp.net c# controller

I changed some things around for an item that was requested. This required me to add a parameter to one of the controller ActionResults.
public ActionResult RejectDocument(int id = 0, string rejectReason)
{
IPACS_Version ipacs_version = db.IPACS_Version.Find(id);
ipacs_version.dateDeleted = System.DateTime.Now;
ipacs_version.deletedBy = User.Identity.Name.Split("\\".ToCharArray())[1];
db.SaveChanges();
return RedirectToAction("Approve");
}
Also this link needs to be activated from jQuery after some jQuery items finish. How do I pass this link off now?
should it be href= Document/RejectDocument?id=222&rejectReason=this is my reject reason
could I then do window.location = href; and it would call the controller and pass in the correct information?

You could use the Url helper to create the right url using the table routes rules. For sample:
window.location = '#Url.Action("RejectDocument", "YourController", new { id = 222, rejectReason = "this is my reject reason" })';

ASP MVC Redirect without changing URL(routing)

Goal:
I want to be able to type URL: www.mysite.com/NewYork OR www.mysite.com/name-of-business
Depending on the string I want to route to different actions without changing the URL.
So far I have:
public static void RegisterRoutes(RouteCollection routes)
{
routes.MapRoute(
"UrlRouter", // Route name
"{query}", // URL with parameters
new { controller = "Routing", action = "TestRouting" } // Parameter defaults
);
}
In the controller I have:
public ActionResult TestRouting(string query)
{
if (query == "NewYork")
return RedirectToAction("Index", "Availability"); // <--------- not sure
else if (query == "name-of-business")
return Redirect("nameofbusines.aspx?id=2731"); // <--------- not sure
else
return RedirectToAction("TestTabs", "Test"); // <--------- not sure
}
I have pretty much tried everything to redirect/transfer to the page without
changing the URL, but everything I've tried changes the URL or gives me an error.
Basically I'm looking for the equivalent of server.transfer where I can keep the URL but send info to the action and have it display its result.

I'm with Nick on this one, though I think you could just use regular views instead of having to do partials. You may need to implement them as shared views if they are not in the views corresponding to the controller (since it will only look in the associated and shared views).
public ActionResult TestRouting(string query)
{
if (query == "NewYork")
{
var model = ...somehow get "New York" model
return View("Index", model );
}
else if (query == "name-of-business")
{
var model = ...get "nameofbusiness" model
return View("Details", model );
}
else
{
return View("TestTabs");
}
}
Each view would then take a particular instance of the model and render it's contents using the model. The URL will not change.
Anytime that you use a RedirectResult, you will actually be sending an HTTP redirect to the browser and that will force a URL change.

Im not sure if you tried this way or if this way has any drawbacks..
Add a global.asax file to your project. In that add the following method:
void Application_BeginRequest(object sender, EventArgs e)
{
// Handles all incoming requests
string strURLrequested = Context.Request.Url.ToString();
GetURLToRedirect objUrlToRedirect = new GetURLToRedirect(strURLrequested);
Context.RewritePath(objUrlToRedirect.RedirectURL);
}
GetURLToRedirect can be a class that has the logic to find the actual URL based on the URL typed in. The [RedirectURL] property will be set with the url to redirect to beneath the sheets.
Hope that helps...

You can change your controller like this:
public ActionResult TestRouting(string query)
{
string controller,action;
if (query == "NewYork")
{
controller = "Availability";
action = "Index";
}
else
{
controller = "Test";
action = "TestTabs";
}
ViewBag.controller = controller;
ViewBag.action = action;
return View();
}
Then you can use these ViewBags in your view like this:
#{
Layout = null;
Html.RenderAction(ViewBag.action, ViewBag.controller);
}
That's it. And you can improve this example with use a class and some functions.

Are you saying you want to go to "www.mysite.com/NewYork" and then "really" go "somewhere else" but leave the url alone? Perhaps what you would want to do then is use partial views to implement this? That way, your base page would be what gets routed to, and then inside of that page you do your condition testing to bring up different partial views? I've done that in my application for viewing either a read-only version of a grid or an editable grid. It worked very nicely.

I'm not sure what you can do about the redirect to the .aspx page, but you should be able to replace the RedirectToAction(...)s with something like this:
public ActionResult TestRouting(string query)
{
if (query == "NewYork")
{
var controller = new AvailabilityController();
return controller.Index();
}
else if (query == "name-of-business")
return Redirect("nameofbusines.aspx?id=2731"); <--------- not sure
else
{
var controller = new TestController();
return controller.TestTabs();
}
}

How can I maintain ModelState with RedirectToAction?

How can I return the result of a different action or move the user to a different action if there is an error in my ModelState without losing my ModelState information?
The scenario is; Delete action accepts a POST from a DELETE form rendered by my Index Action/View. If there is an error in the Delete I want to move the user back to the Index Action/View and show the errors that are stored by the Delete action in the ViewData.ModelState. How can this be done in ASP.NET MVC?
[AcceptVerbs(HttpVerbs.Post | HttpVerbs.Delete)]
public ActionResult Delete([ModelBinder(typeof(RdfUriBinder))] RdfUri graphUri)
{
if (!ModelState.IsValid)
return Index(); //this needs to be replaced with something that works :)
return RedirectToAction("Index");
}

Store your view data in TempData and retrieve it from there in your Index action, if it exists.
...
if (!ModelState.IsValid)
TempData["ViewData"] = ViewData;
RedirectToAction( "Index" );
}
public ActionResult Index()
{
if (TempData["ViewData"] != null)
{
ViewData = (ViewDataDictionary)TempData["ViewData"];
}
...
}
[EDIT] I checked the on-line source for MVC and it appears that the ViewData in the Controller is settable, so it is probably easiest just to transfer all of the ViewData, including the ModelState, to the Index action.

Use Action Filters (PRG pattern) (as easy as using attributes)
Mentioned here and here.

Please note that tvanfosson's solution will not always work, though in most cases it should be just fine.
The problem with that particular solution is that if you already have any ViewData or ModelState you end up overwriting it all with the previous request's state. For example, the new request might have some model state errors related to invalid parameters being passed to the action, but those would end up being hidden because they are overwritten.
Another situation where it might not work as expected is if you had an Action Filter that initialized some ViewData or ModelState errors. Again, they would be overwritten by that code.
We're looking at some solutions for ASP.NET MVC that would allow you to more easily merge the state from the two requests, so stay tuned for that.
Thanks,
Eilon

In case this is useful to anyone I used #bob 's recommended solution using PRG:
see item 13 -> link.
I had the additional issue of messages being passed in the VeiwBag to the View being written and checked / loaded manually from TempData in the controller actions when doing a RedirectToAction("Action"). In an attempt to simplify (and also make it maintainable) I slightly extended this approach to check and store/load other data as well. My action methods looked something like:
[AcceptVerbs(HttpVerbs.Post)]
[ExportModelStateToTempData]
public ActionResult ChangePassword(ProfileViewModel pVM) {
bool result = MyChangePasswordCode(pVM.ChangePasswordViewModel);
if (result) {
ViewBag.Message = "Password change success";
else {
ModelState.AddModelError("ChangePassword", "Some password error");
}
return RedirectToAction("Index");
}
And my Index Action:
[ImportModelStateFromTempData]
public ActionResult Index() {
ProfileViewModel pVM = new ProfileViewModel { //setup }
return View(pVM);
}
The code in the Action Filters:
// Following best practices as listed here for storing / restoring model data:
// http://weblogs.asp.net/rashid/archive/2009/04/01/asp-net-mvc-best-practices-part-1.aspx#prg
public abstract class ModelStateTempDataTransfer : ActionFilterAttribute {
protected static readonly string Key = typeof(ModelStateTempDataTransfer).FullName;
}
:
public class ExportModelStateToTempData : ModelStateTempDataTransfer {
public override void OnActionExecuted(ActionExecutedContext filterContext) {
//Only export when ModelState is not valid
if (!filterContext.Controller.ViewData.ModelState.IsValid) {
//Export if we are redirecting
if ((filterContext.Result is RedirectResult) || (filterContext.Result is RedirectToRouteResult)) {
filterContext.Controller.TempData[Key] = filterContext.Controller.ViewData.ModelState;
}
}
// Added to pull message from ViewBag
if (!string.IsNullOrEmpty(filterContext.Controller.ViewBag.Message)) {
filterContext.Controller.TempData["Message"] = filterContext.Controller.ViewBag.Message;
}
base.OnActionExecuted(filterContext);
}
}
:
public class ImportModelStateFromTempData : ModelStateTempDataTransfer {
public override void OnActionExecuted(ActionExecutedContext filterContext) {
ModelStateDictionary modelState = filterContext.Controller.TempData[Key] as ModelStateDictionary;
if (modelState != null) {
//Only Import if we are viewing
if (filterContext.Result is ViewResult) {
filterContext.Controller.ViewData.ModelState.Merge(modelState);
} else {
//Otherwise remove it.
filterContext.Controller.TempData.Remove(Key);
}
}
// Restore Viewbag message
if (!string.IsNullOrEmpty((string)filterContext.Controller.TempData["Message"])) {
filterContext.Controller.ViewBag.Message = filterContext.Controller.TempData["Message"];
}
base.OnActionExecuted(filterContext);
}
}
I realize my changes here are a pretty obvious extension of what was already being done with the ModelState by the code # the link provided by #bob - but I had to stumble on this thread before I even thought of handling it in this way.

Please don't skewer me for this answer. It is a legitimate suggestion.
Use AJAX
The code for managing ModelState is complicated and (probably?) indicative of other problems in your code.
You can pretty easily roll your own AJAX javascript code. Here is a script I use:
https://gist.github.com/jesslilly/5f646ef29367ad2b0228e1fa76d6bdcc#file-ajaxform
(function ($) {
$(function () {
// For forms marked with data-ajax="#container",
// on submit,
// post the form data via AJAX
// and if #container is specified, replace the #container with the response.
var postAjaxForm = function (event) {
event.preventDefault(); // Prevent the actual submit of the form.
var $this = $(this);
var containerId = $this.attr("data-ajax");
var $container = $(containerId);
var url = $this.attr('action');
console.log("Post ajax form to " + url + " and replace html in " + containerId);
$.ajax({
type: "POST",
url: url,
data: $this.serialize()
})
.done(function (result) {
if ($container) {
$container.html(result);
// re-apply this event since it would have been lost by the form getting recreated above.
var $newForm = $container.find("[data-ajax]");
$newForm.submit(postAjaxForm);
$newForm.trigger("data-ajax-done");
}
})
.fail(function (error) {
alert(error);
});
};
$("[data-ajax]").submit(postAjaxForm);
});
})(jQuery);

Maybe try
return View("Index");
instead of
return Index();