I want to return a JSON result. To do this I have a controller method as follows that is called from a Ajax.BeginForm on the View:
#using (Ajax.BeginForm("Update", new AjaxOptions { OnSuccess = "MySuccessMethod()" }))
{
<!-- some form stuff -->
<input type="submit" value="Submit"/>
}
This is the controller that handles it:
[HttpPost]
public JsonResult Update(FormCollection fc)
{
// Process form stuff
return Json (new {success = true });
}
What I want is to process the success response with MySuccessMethod. What I see is that the view on submit goes to the correct controller method above, which then redirects the page to the URL /Home/Update with the following string in the screen:
{"success": true }
Not sure if it is relevant but I am using Mono.
How can I make the framework not switch pages to /Home/Update nor display the JSON string on the view and just process the JSON in the back?
For your first question, check the following:
1) Make sure you have Microsoft.jQuery.Unobtrusive.Ajax included and referenced
2) OnSuccess = "MySuccessMethod()" should be OnSuccess = "MySuccessMethod" (where MySuccessMethod is a JavaScript method, not a C# one)
For your second question, you could have your method return ActionResult instead of JsonResult (see here for more information). JsonResult is a type of ActionResult, which means that updating your action to return ActionResult will allow your method to return multiple types of ActionResult depending on the scenario:
[HttpPost]
public ActionResult SomeThing(int randomParam)
{
if (randomParam == 0)
{
return Json("Zero!");
}
else if (randomParam == 1)
{
return View("Not zero!");
}
else
{
return HttpNotFound("Error: I can only find zeroes and ones");
}
}
As a general rule of thumb (although sometimes rules are meant to be broken), having your action return one type (like your example, JsonResult instead of ActionResult) makes your action less error-prone as, for example, Visual Studio will let you know if you accidentally try to return another type of result - use ActionResult when your action returns more than one type of result.
Related
I have a Controller with some actions on it as follows:
[HttpPost]
public ActionResult Create(CreateModel model)
{
if (model.SelectedCustomers.Count > 0 &&
model.SelectedVersions.Count > 0 &&
!string.IsNullOrWhiteSpace(model.ScriptName) &&
!string.IsNullOrWhiteSpace(model.ScriptText))
{
Script script;
...save to database...
return Edit(script.Id); //<---------Return other view here
}
else
{
...
}
}
[HttpGet]
public ActionResult Edit(int? scriptId)
{
return View();
}
After the Create action runs, and saves my model to the database successfully, I want to send the user to the Edit view for the newly created script. When I use the code above, specifically return Edit(script.Id); it just sends the user back to the Create view instead of the Edit view. When the user navigates to the Edit action directly, or through the result of an Html.ActionLink pointed at Edit everything works correctly.
What am I doing wrong?
This isn't doing what you think it does:
return Edit(script.Id)
It's not actually telling the framework to go to that action. It's just returning the return value of that method. Purely a C# concern before any components of the ASP.NET MVC Framework are involved at all. And what is that return value:
return View()
So the former is really functionally the same thing as the latter. And any time you use return View() in ASP.NET MVC, the framework will determine that view by examining the action currently being called, which in this case is Create.
What you want isn't to return the Edit view (even if you do, in this case, the user is still on the Create URL, which will cause confusion). What you want is to return a redirect to tell the client to request that next action:
return RedirectToAction("Edit", new { scriptId = script.Id });
You can always call RedirectToAction and return that action result. That will inform the browser to redirect to the different action.
I think you will need something like this:
return RedirectToAction("Edit", new { scriptId = script.Id });
Calling Edit directly is no different than calling a method.
You can do with this RedirecToAction with input parameters.
return RedirectToAction("Action", new { id = 12 });
In Your Case:
return RedirectToAction("Edit", new { scriptId = script.Id });
I have created a page that takes in a string that will search a list of vendors. My goal is to output them to a grid list on an HTML page. Oddly enough, the first page loads, and I can break point the code until the return view of the actual list page. However, it never actually loads. It is even more frustrating because if I don't pass the model to the grid page, it gives me the typical "You can't use a null model", but then it still doesn't load the new page. I have tried several versions. The most current is below.
[HttpPost]
public ActionResult Search(String searchString)
{
this.searchString = searchString;
List<VendorInvoice> v = VendorSearches.publicSearch(searchString);
test = v;
ViewData.Model = v;
TempData.Add("test",v);
return RedirectToAction("Search");
}
[HttpGet]
public ActionResult Search()
{
List<VendorInvoice> v = (List<VendorInvoice>)TempData["test"];
return View("Search",v);
}
So if I take the v out, then I get the error about not passing the model. If it is there, then nothing will happen. The new page won't load.
In your HttpPost search action method, you are setting the result data to show in TempData and doing calling the RedirectToAction method.
RedirectToAction returns an HTTP 302 response to the browser, which causes the browser to make a GET request to the specified action. That means, it will be totally new request coming to your search GET action again. Since Http is stateles, it does not have any idea about what you did in your previous request. The data stored in TempData won't be available to this request.
What you should be doing is, similar to your GET action method, simply
return the result to the view.
[HttpPost]
public ActionResult Search(String searchString)
{
this.searchString = searchString;
List<VendorInvoice> v = VendorSearches.publicSearch(searchString);
return View("Search",v);
}
That should fix your problem. But as Stephen Muecke mentioned, you can keep just your GET action method for your Initial view and search result view
public ActionResult Search(String searchString="")
{
List<VendorInvoice> v = new List<VendorInvoice>();
v = VendorSearches.publicSearch(searchString);
return View("Search",v);
}
And your view
#model List<VendorInvoice>
#using(Html.BeginForm("Search","YourControllerName",FormMethod.GET)
{
<input type="text" name="searchString" />
<input type="submit" />
}
<h2>Results</h2>
#foreach(var item in Model)
{
<p> #item.SomePropertyNameOfYourVendorInvoiceHere </p>
}
I don't have a view called Test. But, there's a method in the Home controller called Test.
Everything works fine, the Test method gets executed and redirected to Index view as given in the code. However, in the browser the URL is ../Home/Test and not ../Home/Index. I don't have a View called Test so why is this getting displayed ? I don't want this URL to be displayed in the browser. How can i solve this ?
View:
#using (Html.BeginForm("Test", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<input type="file" name="file" />
..// Other code
}
C#
public ActionResult Test(HttpPostedFileBase f)
{
var m = new HomeModel();
..// Other code goes here
return View("../Home/Index", m); // Will be returning the Index View
}
In URL
"../Home/Test"
Test is action thus it is working fine.
If you don't want this URL to be displayed, renamed the Action "Test" to "Index" and also update its references
and use (optional)
return View("Index", m)
If you are using the standard setting for MVC, it is not the View, but The Method, that is displayed in the browser. The method then by default returns the View (but that is optional).
So what you need to do is rename you Test Method to Index and place [HttpPost] on top of it.
[HttpPost]
public ActionResult Index(HttpPostedFileBase f)
{
var m = new HomeModel();
..// Other code goes here
return View(m);
}
return View("../Home/Index", m);
This will not redirect you to Index; it will simply display your Index view. If you want browser to automatically change your URL from /Test to /Index you have to instead do this:
return RedirectToAction("Index");
Try this:
public ActionResult Test(HttpPostedFileBase f)
{
var m = new HomeModel();
..// Other code goes here
return RedirectToAction("Index");
}
I have this problem:
I go to a page such as:
/Auction/Details/37
and this calls this action method:
public ActionResult Details(int id)
A particular line in this method is:
return View("DetailsLub", auction);
This view contains this line:
#Html.Action("BidOnAuction", new { auctionId = Model.Id })
Which calls this action method:
public PartialViewResult BidOnAuction(int auctionId)
So far so good?
Now, I have a form in the BidOnAuction view, whcih has a button. When I click on this button, this action method is invloked:
[HttpPost]
public ActionResult BidOnAuction(BidOnAuctionViewModel model)
This action method has a catch statement with the following lines:
ModelState.AddModelError(string.Empty, operation + #" Failure: " + message);
return RedirectToAction("Details", new { id = model.AuctionId });
Now, both the DetailsLUB view and the BidOnAction view contain this line:
#Html.ValidationSummary(true)
But, the issue is that nothing ever gets printed to the screen. What am I doing wrong?
InOrder to get the validation Message on the page you need to return view with Model, as model has the Model State within it, something like this:
return View(Model);
This will return the model BidOnAuction with Validation Summary.
This line of code
return RedirectToAction("Details", new { id = model.AuctionId });
Returns instance of RedirectResult class. That is generally used for redirections and does not render view. If you want to render child action into parent view using #Html.Action, you need to return view from that child action, not RedirectResult. And that RedirectResult will not work even when there's no child action. Returning RedirectResult causes browser to issue fresh, all new request to that action. And model state is lost anyways. You should do something like
try
{
//some actions
return RedirectResult("Details", new { id = model.AuctionId });
}
catch
{
ModelState.AddModelError(string.Empty, operation + #" Failure: " + message);
return View("Details", new { id = model.AuctionId });
}
You can't redirect to a new action and expect the modelstate to be there.
If the modelState is invalid just return (with View(model))
else
redirect to details.
If you need the error information in the details view you will have add it to TempData or pass it in as an optional parameter.
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();