Using a generic model in ASP.NET MVC Razor

Question

Is it possible to use a generic model in ASP.NET MVC 3 (w/ Razor)? The following fails with a syntax error:

@model DtoViewModel<T> where T : IDto

Answer 1

This is not ideal, but it works and could get pretty creative with this pattern.

@model YourNameSpace.MyModel


public MyModel
{
    public MyGenericType<string> ModelAStuff {get;set;}
    public MyGenericType<int> ModelBStuff {get;set;}
    public MyGenericType<DateTime> ModelCStuff {get;set;}
}

public class MyGenericType<T>
{
  //use T how ever you like
  public T Color {get;set;}
  public T Year  {get;set;}
}

Answer 2

given that @model is expecting a type - not a type declaration you could use:

@model DtoViewModel<IDto>

and take advantage of generic covariance

Answer 3

I found this thread and did some further digging. Looks like the feature is on the 2.2.0 backlog. If anyone wants to get involved you can check out the issue on Github. https://github.com/aspnet/Mvc/issues/7152

Answer 4

Such syntax is not supported by Razor, sorry.

Answer 5

Assuming, you want to use a generic type in order to avoid code duplications in each view of ViewModel<T> you can do it this way:

1. create a view for the parts of ViewModel<T> that are unique to the view

ModelView.cshtml:

@model ViewModel<specificType>

@{Layout = "~/Views/Shared/Layout.cshtml";}
<h2 class="sub-header">Specific type view</h2>

2. create a view for the common parts, that should be rendered in each view of <T>

Grid.cshtml:

@{ var webGrid = new WebGrid(Model.PageItems); }

<div class="row" style="overflow: auto">
    @webGrid.GetHtml("table-striped", mode: WebGridPagerModes.All, firstText: "First", lastText: "Last")
</div>

Since it's a partial view, you don't need to declare the type of Model again. It will simply use the model you defined in the parent view, that renders it. The property IList<T> PageItems of your model, will remain strongly typed with <specificType>.

3. Don't forget to actually render the partial view of your common parts

ModelView.cshtml:

@RenderPage("~/Views/Shared/Grid.cshtml")