How can I maintain ModelState with RedirectToAction?

spoon16_3957 0
Viewed 41539

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");
}
asp.net-mvc c#
6 Answers

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.

user_129500

Use Action Filters (PRG pattern) (as easy as using attributes)

Mentioned here and here.

bob_238050

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

Eilon_316680

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);
user_18046780

Maybe try

return View("Index");

instead of

return Index();
Ty_169480
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