CODEWITHSUNDEEP
c#asp.netasp.net-mvcasp.net-mvc-5
Martin
Martin

Reputation: 320

ASP Net MVC - Send different model on POST

is it possible to send an object from a strongly typed view to the Controller via Http-POST that does not equal the type of the original model.

For example:

I have a ViewModel like this:

public class PersonsViewModel
{
    List<PersonViewModel> persons { get; set; }

    PersonsViewModel() { }
}

public class PersonViewModel
{
    //some properties

    Person() { }
}

Now i have this View:

@model PersonsViewModel

<div>
  @for(int i = 0; i > Model.persons.Count; i++)
{
    @Html.EditorFor(Model.persons[i])
}
</div>

The editor could look like this:

@model PersonViewModel

<div>
     @using (Html.Beginform("Postaction","Controller", FormMethod.Post)){
     <div>
       <!-- properties and textboxes here + submit button -->
     </div>
}
<div>

The controller action

[ValidateAntiForgeryToken]
        [HttpPost]
        public ActionResult Postaction(PersonViewModel model)
        {
            //do something
        }
    }

This doesn't work because it seems the Controller is expecting a PersonsViewModel object. My workaround so far is to make a "big" Form that contains all PersonViewModel and send the complete PersonsViewModel to the controller.

Is it somehow possible to pass only one PersonViewModel to the Controller although the view is strongly typed?

Kind regards, Martin

Upvotes: 4

Views: 7295

Answers (1)

Alex Art.
Alex Art.

Reputation: 8781

It could be done:

When used with collections Html.EditorFor is smart enough to generate input names that contain index so ModelBinder could successfully create a model as a collection of objects. In your case since you want to have a separate form per PersonViewModel object, you could create a partial view as a template for editing PersonViewModel and use Html.RenderPartial helper:

Assuming you have _PersonViewModel.cshtml partial view 

@for(int i = 0; i > Model.persons.Count; i++)
{
    Html.RenderPartial("_PersonViewModel", Model.persons[i]);
}

in the _PersonViewModel.cshtml you can not use neither one of editor helpers such as Html.EditorFor, Html.TextboxFor because they are going to generate identical ids for the same properties so you will have to manually create html inputs:

@model PersonViewModel

<div>
     @using (Html.Beginform("Postaction","Controller", FormMethod.Post)){
     <div>
       @*Nottice the usage of Html.NameFor(m=>m.FirstName) for generating a name property value *@
       <input type="text" name="@Html.NameFor(m=>m.FirstName)" value="@Model.FirstName"> 
     </div>
}
<div>

This way you can post a single PersonViewModel object to the controller action

Upvotes: 6

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