I created a custom Model i.e. to support my Razor View. Then I created a controller as following`namespace MyCandidate.Controllers
public class CandidateViewModelController : Controller
{
//
// GET: /CandidateViewModel/
public ActionResult Index()
{
return View();
}
}
I also have the following statement in my _Layout.cshtml
#Html.ActionLink("Canid", "Index", "CandidateViewModel")
Next i created a view and the very first statement of the view is
#model MyCandidate.Models.CandidateViewModel
when i run my project i get the following error
The view 'Index' or its master was not found or no view engine supports the searched locations. The following locations were searched:
I spent more than 3 hours but could not figure out?
your Index() not get any parameters, but you send "CandidateViewModel") add Index(string input) method with attribute [HttpGet] to controller.
this error meen you haven't view "Index" in "Views/CandidateViewModel/Index.cshtml".
Maybe you delete master page files (_ViewSrat, _Layout)
Or you made a mistake when you change your routes
Change your ActionLink to:
#Html.ActionLink("Canid", "Index")
If you want to pass any data to View , you can also use ViewBag:
// Controller :
ViewBag.CandidateValues = CandidateViewModelData;
// View
#Html.Label("LabelName", (CandidateViewModel) ViewBag.CandidateValues.FiledName);
Related
I use a child action in Home Controller like this:
[OutputCache(Duration=30)]
public ActionResult ChildAction()
{
Response.Write("Date Time is="+DateTime.Now);
return View();
}
Home.cshtml
<h2>About</h2>
<p>
#Html.Action("ChildAction")
</p>
but after run my asp.net mvc3.0 Project i get an error :
The view 'ChildAction' or its master was not found or no view engine supports the searched locations. The following locations were searched:
~/Views/Home/ChildAction.aspx
~/Views/Home/ChildAction.ascx
~/Views/Shared/ChildAction.aspx
~/Views/Shared/ChildAction.ascx
~/Views/Home/ChildAction.cshtml
~/Views/Home/ChildAction.vbhtml
~/Views/Shared/ChildAction.cshtml
~/Views/Shared/ChildAction.vbhtml
please help me how can i solve it
thanks
You should have a view named "ChildAction", or in action you should return view as
return View("ViewName"); In summary, view name should be same as action name, not controller name as you did(You named view as "Home").
I currently have a _layout.cshtml used by every page of my website.
I need to put a form on each page displayed as a popin.
So, i created a new PartialView (the content of my form) with its corresponding ViewModel and called it in _layout.cshtml.
However, i have a model conflict between ViewModels of pages using the layout and the ViewModel used by the new form (since we can't have directly two models for the same view).
The model item passed into the dictionary is of type 'XXX', but this
dictionary requires a model item of type 'YYY'.
How can we include a form in _layout without this conflict ?
The following has worked for me with a sidebar on every page.
Create a controller for your partial view
In that controller, create a method for the view you want to return, and be sure to use the [ChildActionOnly] filter
public class PartialController : Controller
{
[ChildActionOnly]
public PartialViewResult Alerts()
{
return PartialView("Alerts", messages);
}
}
In your _layout view, you'll have the following:
#Html.Action("Alerts", "Partial")
(instead of #Html.RenderPartial or #Html.Partial)
It sounds like you already have what you need for the view.
I have not used this with a form, but it should work similarly. Hope this helps.
I have been playing around with Umbraco 5 for some days now. I have made a partial view with some dummy-text that I have inserted into a page template. Work's fine. The problem is when I have to pass data from a controller to the view.
The view inherit from RenderViewPage (#inherits RenderViewPage) as default in Umbraco 5. I tried to do it the regular MVC way by #model ViewPage<Umbraco.Cms.Web.UI.Models.Test> but I got an error.
You should start by creating a Surface controller (can be done in seperate project or create controller folder directly in main project):
public class ContactFormSurfaceController : SurfaceController
{
[ChildActionOnly]
public PartialViewResult ContactForm()
{
var model = new ContactViewModel();
return PartialView(model);
}
}
Don't inherit form RenderViewPage just strongly type your view with your own model
Then create a macro that call the ChildAction ContactForm
You can add your action through editor or via code in templates: #Umbraco.RenderMacro("ContactForm")
I'm looking into using partial views in MVC3 using Razor, and I get my partial view to render and it works fine.
What I'd like to do, though, is refresh the parent view when the partial view is submitted.
Code in my parent view to render partial view
<div id="mydiv">
#{ Html.RenderAction("Add", "Request"); }
</div>
Action for parent view is simple,
public ActionResult Index()
{
List<obj> reqs = //some query
return View(reqs);
}
In my partial view's get action I have:
public ActionResult Add()
{
AddRequestViewModel vm = new AddRequestViewModel();
//set some stuff on the VM here
return PartialView(vm);
}
In the post action called by the partial view, if modelstate isn't valid, return PartialView(vm)
If it is valid, I'd like the parent and partial views to refresh.
I tried RedirectToAction, but this can't be called in an action called by a partial, apparently, and I tried return Index();, but this causes an issue with the code used to render the partial view,
Exception Details: System.InvalidOperationException: The model item passed into the dictionary is of type 'System.Collections.Generic.List'1[DatRequests.Models.ReqRequest]', but this dictionary requires a model item of type 'DatRequests.ViewModels.AddRequestViewModel'.
Any suggestions on how to do this would be appreciated. The purpose of the page is to show a list of elements, and the partial contains a form to add a new element to the list.
Edit: The partial's model is different, as it contains data for selection, which is from a db, which is why I tried RenderAction, but I'm not sure if there are other ways of doing this.
When the partial view is submitted normally you submit it to some controller action. You could either submit it using a normal request or an AJAX request. If you use a normal request you could perform a standard redirect to the Index inside the POST controller action that will handle the form submission. If you use AJAX, you could return a JSON result pointing to the url that you want to redirect:
[HttpPost]
public ActionResult Foo(MyViewModel model)
{
if (!ModelState.IsValid)
{
return PartialView(model);
}
return Json(new { url = Url.Action("Index") });
}
and inside your AJAX success callback:
success: function(result) {
if (result.url) {
// we have a success
window.location.href = result.url;
} else {
// invalid modelstate => refresh the partial
$('#mydiv').html(result);
}
}
Probably RenderAction should not be used this way.
When using Html.RenderAction, a new/seperate request would be sent to the server. And you got another chance to load some data from db or somewhere else to display to the client. Also, you could apply OutputCache to this action. this is usually the way doing global cache.
Here you are doing a POST to the server. Either directly put a element here or using a partial view to do the Post. And in the corresponding action, do a RedirectToAction.
Do it with ajax or not isn't the point. my opinion is more about the right way using RenderAction
I am using ASP.NET MVC 3 for my website.
I have created a partial view that has "Previous, Next and Save" buttons. I am calling this partial view on my master page.
My requirement is that on what ever View I am I must be able to call different Save methods in different controllers by passing respective Model data to controller actions.
Example
I have 4 step data input, I have a different controller for each step.
If i am on step 1 and i click Save the form Values should go to Step1Controller's action method,
If I am on step 2 then post should call Step2Controller
Something like this this:
public ActionResult Save(GenericModel model)
{
//use reflection to find out model type
//call appropriate controller action with model
return RedirectToAction("Create", new { Controller = "Conference", Action = "Create" });
}
This save method will be called for Save button on the Master page. How can I achieve this?
Are the forms on separate actions on the controllers?
If so, just set the form action on each page to point to the relevant controller. So form 1 is
<form method="post" action="/step1controller/action">
Form 2 is:
<form method="post" action="/step2controller/action">
Does that solve your problem?
I would create a private method in the controller and call in every part of the first steps.
private bool Save(GenericModel model)
{
......
}
[HttpPost]
public bool SaveStep1(GenericModel model)
{
this.Save(model);
}
[HttpPost]
public bool SaveStep2(GenericModel model)
{
this.Save(model);
}