Custom ActionResult equivalent Model for all PartialViews

Question

I created following custom ActionResult which returns multiple partial views.

public class MultiplePartialViewsResult : ActionResult
{
    private const string Separator = "-";
    private PartialViewResult[] _partialViews;

    public MultiplePartialViewsResult(params PartialViewResult[] partialViews)
    {
        _partialViews = partialViews;
    }

    public override void ExecuteResult(ControllerContext context)
    {
        foreach (var partialView in _partialViews)
        {
            partialView.ExecuteResult(context);
            context.HttpContext.Response.Output.Write(Separator);
        }
    }
}

And then I use it as follows:

return new MultiplePartialViewsResult(
             PartialView("~/Views/RowSumView.cshtml", new List<double>() { 1.0 }),
             PartialView("~/Views/ColumnSumView.cshtml", new List<double>() { 2.0 }));

But if I place breakpoint in MultiplePartialViewsResult's constructor I see that model in both cases equals 2.0. It overrides all models with the last model specified in a array.

Try to create couple PartialViews in method and assign it to variable. Then you will notice they all share equivalent model.

Yes it is correct.If you are trying to return two `model`s to one view, create a `view model` that contains both of the models that you want to send, and make your view's model the new `ViewModel` — Parth Trivedi, Jan 10 '16 at 14:19
That is the point that I am trying to return to many views.Idea is based on this post where author indicates that models can be distinct. https://www.simple-talk.com/dotnet/asp.net/revisiting-partial-view-rendering-in-asp.net-mvc/ — Maximus, Jan 10 '16 at 14:28
Have check last things? It says to use `view model` shared by both `partial view`. — Parth Trivedi, Jan 10 '16 at 14:31
"In the example, the two partial views use the same view model. This is not a limitation though. " — Maximus, Jan 10 '16 at 14:34
Why would I if an author states it is not obligatory. Apparently something under the cover acts in PartialView that it saves only last model used. — Maximus, Jan 10 '16 at 14:38
May be. You have to wait for better answer to judge this issue. — Parth Trivedi, Jan 10 '16 at 14:45

Mikhail · Answer 1 · 2016-01-26T08:11:26.800

You can initialize the MultiplePartialViewsResult object in the following manner with new own ViewData / ViewData.Model instances (works fine for me):

return new MultiplePartialViewsResult(
    //PartialView("~/Views/RowSumView.cshtml", new List<double>() { 1.0 }),
    //PartialView("~/Views/ColumnSumView.cshtml", new List<double>() { 2.0 })
    new PartialViewResult() { ViewName = "~/Views/RowSumView.cshtml", ViewData = new ViewDataDictionary() { Model = new List<double>() { 1.0 } } },
    new PartialViewResult() { ViewName = "~/Views/ColumnSumView.cshtml", ViewData = new ViewDataDictionary() { Model = new List<double>() { 2.0 } } }
);

Custom Action Result:

public class MultiplePartialViewsResult : ActionResult {
    private const string Separator = "-";
    private PartialViewResult[] _partialViews;

    public MultiplePartialViewsResult(params PartialViewResult[] partialViews) {
        _partialViews = partialViews;
    }

    public override void ExecuteResult(ControllerContext context) {
        foreach(var partialView in _partialViews) {
            partialView.ExecuteResult(context);
            context.HttpContext.Response.Output.Write(Separator);
        }
    }
}

Controller:

public ActionResult Index() {
    return View();
}
public ActionResult FakeAction() {
    return new MultiplePartialViewsResult(
        //PartialView("~/Views/RowSumView.cshtml", new List<double>() { 1.0 }),
        //PartialView("~/Views/ColumnSumView.cshtml", new List<double>() { 2.0 })
        new PartialViewResult() { ViewName = "~/Views/RowSumView.cshtml", ViewData = new ViewDataDictionary() { Model = new List<double>() { 1.0 } } },
        new PartialViewResult() { ViewName = "~/Views/ColumnSumView.cshtml", ViewData = new ViewDataDictionary() { Model = new List<double>() { 2.0 } } }
    );
}

Views:

Index:

@Html.Action("FakeAction")

ColumnSumView.cshtml / RowSumView.cshtml:

@model List<double>
<ul>
    @foreach(double item in Model) {
        <li>@item</li>
    }
</ul>

That's a workaround but I am wondering why my example is not working. Is there a better approach to pass multiple partial views? — Maximus, Jan 26 '16 at 10:55
It seems that it is related to holding a ViewData/Model object while rendering a PartialView (a single ViewData per an ActionResult). Possibly, this is expected behavior. I will try to check if there is a better solution. For now, you can use my suggested approach and always initialize each ActionResult/PartialViewResult with a new ViewData/Model instance. — Mikhail, Jan 26 '16 at 12:47
This happens because of the [`PartialView`](https://aspnetwebstack.codeplex.com/SourceControl/latest#src/System.Web.Mvc/Controller.cs) helper of the controller that Maximum used to create each partial result. That helper will set the model in ViewData.Model where ViewData is a property in [the controller](https://aspnetwebstack.codeplex.com/SourceControl/latest#src/System.Web.Mvc/ControllerBase.cs). Calling twice `PartialResult` will override the ViewData.Model — Daniel J.G., Jan 26 '16 at 13:13

Custom ActionResult equivalent Model for all PartialViews

1 Answers1