I have a view (Index.cshtml) with a submit button. When the submit button is clicked, it calls an action (Action01) within the controller (TestController.cs) so at the end of the action I want to return to the caller (Index.cshtml) view with a custom view model as a parameter. How do I do this?
Results after first attempt using View("ViewName",model):
An error is raised, as the action is within the controller Test, so returning, it is searching for \Views\Tests\Index, and my Index page is in \Views\Home\Index.
The view 'Index' or its master was not found or no view engine supports the searched locations. The following locations were searched:
~/Views/Test/Index.aspx
~/Views/Test/Index.ascx
~/Views/Shared/Index.aspx
~/Views/Shared/Index.ascx
~/Views/Test/Index.cshtml
~/Views/Test/Index.vbhtml
~/Views/Shared/Index.cshtml
~/Views/Shared/Index.vbhtml
I have used return View("ViewName", model)
, and I have changed my directories structure as it was the problem.
This question is related to
asp.net-mvc
asp.net-mvc-4
To return a different view, you can specify the name
of the view you want to return and model
as follows:
return View("ViewName", yourModel);
if the view is in different folder under Views folder then use below absolute path:
return View("~/Views/FolderName/ViewName.aspx");
public ActionResult Index()
{
return View();
}
public ActionResult Test(string Name)
{
return RedirectToAction("Index");
}
Return View
Directly displays your view
but
Redirect ToAction
Action is performed
You have to specify the name of the custom view and its related model in Controller Action method.
public ActionResult About()
{
return View("NameOfViewYouWantToReturn",Model);
}
Also, you can just set the ViewName:
return View("ViewName");
Full controller example:
public ActionResult SomeAction() {
if (condition)
{
return View("CustomView");
}else{
return View();
}
}
This works on MVC 5.
Source: Stackoverflow.com