If I have a view that has a model, lets say Car..
@model Project.Car
inside that view I want to create a form that sends data to a new model
@using (Html.BeginForm("Add", "Controller"))
{
@Html.Hidden("ID", "1")
@Html.Hidden("UserID", "44")
@Html.TextArea("Description")
}
I've noticed that if my action is defined with my ViewModel it does not work (model is always null):
[HttpPost]
public PartialViewResult Add(ViewModels.NewModel model)
However, if I use a FormCollection it works:
[HttpPost]
public PartialViewResult Add(FormCollection formCollection)
Here is the ViewModel:
public class NewModel
{
public int ID { get; set; }
public int UserID { get; set; }
public string Description { get; set; }
}
My question is can I post data to NewModel from my form? The View that it sits on is correct to be tied to Project.Car. Its a small form on the page that needs to post a different set of data that has nothing to do with Project.Car.
You can also create a custom model binder for this purpose.
The short answer is yes you can post the form to any controller action on any controller related to any Model in your application.
For example, for your form to post to the "
Add
" action on theNewModel
controller:Since your view is strongly typed to your
Car
model, You can either change this and send a ViewModel to your view whose type matches the model your updating (as Darin demonstrated), or you'll need to map the post data fromCar
ontoNewModel
in your controller:On the
CarController
'sAdd
action (Post) :Also, check out AutoMapper, an object-to-object mapper which automates mapping of ViewModels onto Models and vice versa.
Yes, you can strongly type a view to one model and POST it to another model.
In doing so you have two options:
Manually provide correct names for each input field, so that the default binder will understand it and create the model (example).
While this works, it also means you have to keep an eye on typos and you will not get any compile-time errors if you misspell a property name.
Manually create a HTML helper in the view bound to the new model type. It will then properly generate the HTML for you.
In order to construct the helper, you need a wrapper object that would expose the instance of your model in the form of the
IViewDataContainer
interface. You can define that wrapper anywhere, including the model itself:Then in a view you create a helper bound to an instance of
NewModel
:And then you use the helper as usual:
Then your
PartialViewResult Add(ViewModels.NewModel model)
will properly receive the data.Your view is set up to use a model of type
Project.Car
however your action method takes a model of typeViewModels.NewModel
, but also your model class posted is of typeAdd
?Change them all to match (assuming
Add
is correct):View:
Controller:
You have a discrepancy between the name of your model and your action. In the example you have shown the model is called
Add
whereas in your action you are usingViewModels.NewModel
. Even worse, your view is strongly typed to a model calledCar
. Messy all this.So start by defining a correct view model:
and then a controller:
and a corresponding strongly typed view to your view model: