Create controller for partial view in ASP.NET MVC

问题:

How can I create an individual controller and model for a partial view? I want to be able to place this partial view any where on the site so it needs it's own controller. I am current rendering the partial as so

@Html.Partial("_Testimonials")

回答1:

Why not use Html.RenderAction()?

Then you could put the following into any controller (even creating a new controller for it):

[ChildActionOnly]
public ActionResult MyActionThatGeneratesAPartial(string parameter1)
{
    var model = repository.GetThingByParameter(parameter1);
    var partialViewModel = new PartialViewModel(model);
    return PartialView(partialViewModel); 
}

Then you could create a new partial view and have your PartialViewModel be what it inherits from.

For Razor, the code block in the view would look like this:

@{ Html.RenderAction("Index", "Home"); }

For the WebFormsViewEngine, it would look like this:

<% Html.RenderAction("Index", "Home"); %>

回答2:

If it were me, I would simply create a new Controller with a Single Action and then use RenderAction in place of Partial:

// Assuming the controller is named NewController
@{Html.RenderAction("ActionName", 
                     "New", 
                      new { routeValueOne = "SomeValue" });
}

回答3:

It does not need its own controller. You can use

@Html.Partial("../ControllerName/_Testimonials.cshtml")

This allows you to render the partial from any page. Just make sure the relative path is correct.

回答4:

The most important thing is, the action created must return partial view, see below.

public ActionResult _YourPartialViewSection()
{
    return PartialView();
}

回答5:

You don't need a controller and when using .Net 5 (MVC 6) you can render the partial view async

@await Html.PartialAsync("_LoginPartial")

or

@{await Html.RenderPartialAsync("PartialName");}

or if you are using .net core 2.1 > you can just use:

<partial name="Shared/_ProductPartial.cshtml"
         for="Product" />

回答6:

Html.Action is a poorly designed technology. Because in your page Controller you can't receive the results of computation in your Partial Controller. Data flow is only Page Controller => Partial Controller.

To be closer to WebForm UserControl (*.ascx) you need to:

1. Create a page Model and a Partial Model

2. Place your Partial Model as a property in your page Model

3. In page's View use Html.EditorFor(m => m.MyPartialModel)
4. Create an appropriate Partial View
5. Create a class very similar to that Child Action Controller described here in answers many times. But it will be just a class (inherited from Object rather than from Controller). Let's name it as MyControllerPartial. MyControllerPartial will know only about Partial Model.
6. Use your MyControllerPartial in your page controller. Pass model.MyPartialModel to MyControllerPartial
7. Take care about proper prefix in your MyControllerPartial. Fox example: ModelState.AddError("MyPartialModel." + "SomeFieldName", "Error")
8. In MyControllerPartial you can make validation and implement other logics related to this Partial Model

In this situation you can use it like:

public class MyController : Controller
{
    ....
    public MyController()
    {
    MyChildController = new MyControllerPartial(this.ViewData);
    }

    [HttpPost]
    public ActionResult Index(MyPageViewModel model)
    {
    ...
    int childResult = MyChildController.ProcessSomething(model.MyPartialModel);
    ...
    }
}

P.S. In step 3 you can use Html.Partial("PartialViewName", Model.MyPartialModel, <clone_ViewData_with_prefix_MyPartialModel>). For more details see ASP.NET MVC partial views: input name prefixes