Is there a way to tell what view a controller action is being called from?
For example, I would like to use "ControllerContext.HttpContext.Request.PhysicalPath" but it returns the path in which the controller action itself is located:
public ActionResult HandleCreateCustomer()
{
// Set up the customer
//..code here to setup the customer
//Check to see of the calling view is the BillingShipping view
if(ControllerContext.HttpContext.Request.PhysicalPath.Equals("~/Order/BillingShipping"))
{
//
return RedirectToAction("OrderReview", "Order", new { id = customerId });
}
else
{
return RedirectToAction("Index", "Home", new { id = customerId });
}
}
If you have a fixed number of locations that it could possibly be called from, you could create an enum where each of the values would correspond to a place where it could have been called from. You'd then just need to pass this enum value into HandleCreateCustomer, and do your condition statement(s) based on that.
At the moment I am using something of the sort:
In the View I am populating a TempData variable using:
#{TempData["ViewPath"] = #Html.ViewVirtualPath()}
The HtmlHelper method ViewVirtualPath() is found in the System.Web.Mvc.Html namespace (as usual) and is as follows and returns a string representing the View's virtual path:
public static string ViewVirtualPath(this HtmlHelper htmlHelper)
{
try{
return ((System.Web.WebPages.WebPageBase)(htmlHelper.ViewDataContainer)).VirtualPath;
}catch(Exception){
return "";
}
}
I will then obviously read the TempData variable in the controller.
I found another way.
In the controller you want to know what page it was called from.
I added the following in my controller
ViewBag.ReturnUrl = Request.UrlReferrer.AbsolutePath;
Then in the View I have a 'Back' button
#(Html.Kendo().Button().Name("ReturnButton")
.Content("Back to List").Events(e => e.Click("onReturn"))
.HtmlAttributes(new { type = "k-button" })
)
Then the javascript for the onReturn handler
function onReturn(e) {
var url = '#(ViewBag.ReturnUrl)';
window.location.href = url;
}
Related
I'm using the following pattern https://github.com/filamentgroup/Ajax-Include-Pattern
to load partial views through ajax.
View:
#using(Html.BeginUmbracoForm("PostContactInformation", "JoiningSurface", null, new Dictionary<string, object> { { "class", "joinform" } })) {
#Html.AntiForgeryToken()
<div data-append="#Url.Action("RenderJoiningContactInformation", "JoiningSurface", new { ContentId = CurrentPage.Id })"></div>
}
With Action:
public ActionResult RenderContactInformation(int ContentId)
{
var viewModel = ContactViewModel();
viewModel.Content = Umbraco.TypedContent(ContentId);
return PartialView("RenderContactInformation", viewModel);
}
Loads partial view perfectly.
// No need to add partial view i think
Post action works correctly as well:
public ActionResult PostContactInformation(ContactViewModel model)
{
//code here
return RedirectToUmbracoPage(pageid);
}
The problem is, that i need to add model error to CurrentUmbracoPage if it exists in post...
For example:
public ActionResult PostContactInformation(ContactViewModel model)
{
ModelState.AddModelError(string.Empty, "Error occurred");
return CurrentUmbracoPage();
}
In this case i get null values for current model. And this happens only when i use ajax.
If i load action synchronously like that:
#using(Html.BeginUmbracoForm("PostJoiningContactInformation", "JoiningSurface", null, new Dictionary<string, object> { { "class", "joinform" } })) {
#Html.AntiForgeryToken()
#Html.Action("RenderContactInformation", "JoiningSurface", new { ContentId = CurrentPage.Id })
}
everything works like it should.
But i need to use ajax. Is there a correct way to pass values on postback in this case? I know that i can use TempData, but i'm not sure that this is the best approach.
Thanks for your patience
The problem is that Umbraco context is not accessible when you're trying to reach it through ajax call. Those calls are a little bit different.
Check my answer in this thread: Umbraco route definition-ajax form and I suggest to go with WebAPI and UmbracoApiControllers to be able to access those values during the Ajax call.
I would like to create a list with a string and an int value at the same time like follows:
#Html.ActionLink("Back to List", "IndexEvent", new { location = "location" })
and
#Html.ActionLink("Back to List", "IndexEvent", new { locationID = 1 })
It didn't work. I guess MVC controller didn't get the type difference of parameter. So, I had to make a new Action as "IndexEvenyByID" but it requires to have a new view. Since I wanted to keep it simple, is there any way to use same view with respect to different parameters?
Try adding two optional parameters to the IndexEvent action like this:
public ActionResult IndexEvent(string location = "", int? locationID = null)
This should not require a new view or view model. You should have two actions as you have described, but the code could be as follows:
Controller
public ActionResult GetEvents(string location){
var model = service.GetEventsByLocation(location);
return View("Events", model);
}
public ActionResult GetEventsById(int id){
var model = service.GetEventsById(id);
return View("Events", model);
}
Service
public MyViewModel GetEventsByLocation(string location){
//do stuff to populate a view model of type MyViewModel using a string
}
public MyViewModel GetEventsById(int id){
//do stuff to populate a view model of type MyViewModel using an id
}
Basically, if your View is going to use the same view model and the only thing that is changing is how you get that data, you can completely reuse the View.
If you really want to stick to a single action and multiple type, you could use a object parameter.
public ActionResult GetEvents(object location)
{
int locationID;
if(int.TryParse(location, out locationID))
var model = service.GetEventsByID(locationID);
else
var model = service.GetEventsByLocation(location as string);
return View("Events", model);
}
Something like that (Not completly right but it gives you an idea). This, however, wouldn't really be a "clean" way to do it IMO.
(Edit)
But the 2 actions method is still by far preferable (eg. What happens if we're able to parse a location name into a int?)
I have a base Controller like follow
public abstract class BaseController
{
protected ActionResult LogOn(LogOnViewModel viewModel)
{
SaveTestCookie();
var returnUrl = "";
if (HttpContext != null && HttpContext.Request != null && HttpContext.Request.UrlReferrer != null)
{
returnUrl = HttpContext.Request.UrlReferrer.LocalPath;
}
TempData["LogOnViewModel"] = viewModel;
return RedirectToAction("ProceedLogOn", new { returnUrl });
}
public ActionResult ProceedLogOn(string returnUrl)
{
if (CookiesEnabled() == false)
{
return RedirectToAction("logon", "Account", new { area = "", returnUrl, actionType, cookiesEnabled = false });
}
var viewModel = TempData["LogOnViewModel"] as LogOnViewModel;
if (viewModel == null)
{
throw new NullReferenceException("LogOnViewModel is not found in tempdata");
}
//Do something
//the problem is I missed the values which are set in the ViewBag
}
}
and another Controller
public class MyController : BaseController
{
[HttpPost]
public ActionResult LogOn(LogOnViewModel viewModel)
{
// base.LogOn is used in differnet controller so I saved some details in view bag
ViewBag.Action = "LogonFromToolbar";
ViewBag.ExtraData = "extra data related only for this action";
return base.LogOn(viewModel);
}
}
the problem is I missed the view bag values in ProceedLogOn action method.
I have the values in Logon method in BaseController.
How can I copy the values of ViewBag from one Action to another Action?
So I can not simply say this.ViewBag=ViewBag;
because ViewBag doesn't have setter. I was thinking of Iterating through viewbag.
I tried ViewBag.GetType().GetFields() and ViewBag.GetType().GetProperties() but they return nothing.
ViewData reflects ViewBag
You can iterate the values you've stored like this :
ViewBag.Message = "Welcome to ASP.NET MVC!";
ViewBag.Answer = 42;
foreach (KeyValuePair<string, object> item in ViewData)
{
// if (item.Key = "Answer") ...
}
This link should also be useful
I'm afraid I don't have the answer how to copy ViewBag.
However, I would never use ViewBag that way.
ViewBag is some data the Controller gives to the View to render output if someone does not like to use ViewModel for some reasons. The View should never know anything about the Controller but your ViewBag is holding a ActionName ;).
Anyway, the ProceedLogOn action method has pretty much parameters which is ... not a nice code actually so why hesitate to add more parameters which are currently being hold in MyController.Logon ViewBag? Then inside method ProceedLogOn you have what you want.
;)
Actually I'm very new to ASP.NET MVC and I need your help.
Here I have some Create Method that takes an argument from the URL to use it as id:
in 'vote' controller :
public ActionResult Create(int id)
{
Meeting meeting = db.Meetings.Find(id); // get the object
ViewBag.meetingID = meeting.meetingID; // get its id and assign it to a ViewBag
return View();
}
and I would like to do something like :
vote.meetingID = #ViewBag.meetingID
in the model so that is directly assgin this property without excplicitely typing it from the HTML view (I mean #Html.EditorFor(model=>meetingID) )
The question is not that clear, but try this.
You could pass the entire model to the view.
Controller:
public ActionResult Create(int id)
{
Meeting meeting = db.Meetings.Find(id); // get the object
ViewData["myMeeting"] = meeting;
return View();
}
To use in the view you can declare it as a variable at the top of the view:
Declare:
#
{
var meetingData = ViewData["myMeeting"] as Meeting;
}
Usage:
<div>
#meetingData.meetingID
</div>
I have a partial view in which there is a form. I POST this form using the PRG pattern. I am using the AjaxHelper to create my form. I also need this form to work without javascript. The problem is that when model validation fails, it always changes the url to my partial view.
public ActionResult PostForm(PostFormModel postFormModel)
{
if (ModelState.IsValid)
{
return RedirectToAction("SomewhereElse");
}
else
{
if (Request.IsAjaxRequest())
{
return PartialView("_PostForm")
}
else
{
// What do I do here?
}
}
}
Here's what I have tried:
return PartialView("_PostForm", postFormModel);
This just renders the partial view and doesn't contain any of the parent stuff.
return View("Index", new ParentModel() { PostFormModel = postFormModel });
This actually produces the correct result. It displays the parent view, but the URL is that of the partial http://localhost:22485/Controller/PostForm! I feel like this is really close to the solution. What now?
If you want to change url, you should redirect to another action (using PRG pattern). Insert next code instead of '// What do I do here?':
postModelService.Save(postFormModel); //to Session or to DB
return RedirectToAction("Parent");
New action should look like this:
public ActionResult Parent()
{
var postFormModel = postModelService.Load();
return View("Index", new ParentModel() { PostFormModel = postFormModel });
}
Hope it helps.