Render partial view with dynamic model in Razor view engine and ASP.NET MVC 3

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When I try to render a partial view whose model type is specified as:

@model dynamic

by using the following code:

@{Html.RenderPartial("PartialView", Model.UserProfile);}

I get the following exception:

'System.Web.Mvc.HtmlHelper<dynamic>' has no applicable method named 'RenderPartial' but appears to have an extension method by that name. Extension methods cannot be dynamically dispatched. Consider casting the dynamic arguments or calling the extension method without the extension method syntax.

However, the same code in a .aspx file works flawlessly. Any thoughts?

Just found the answer, it appears that the view where I was placing the RenderPartial code had a dynamic model, and thus, MVC couldn't choose the correct method to use. Casting the model in the RenderPartial call to the correct type fixed the issue.

source: Using Html.RenderPartial() in ascx files

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Instead of casting the model in the RenderPartial call, and since you're using razor, you can modify the first line in your view from

@model dynamic

to

@model YourNamespace.YourModelType

This has the advantage of working on every @Html.Partial call you have in the view, and also gives you intellisense for the properties.

How to: Render ASP.NET MVC Razor Views to Strings, [SOLVED] Render partial view with dynamic model in Razor view engine and ASP.NET MVC 3 | asp.net-mvc Knowledge Base. Questions: When I try to render a partial view whose model type is specified as: @model dynamic by using the following code: @{Html.RenderPartial("PartialView", Model.UserProfile);} I get the following exception: 'System.Web.Mvc.HtmlHelper<dynamic>' has no applicable method named 'RenderPartial' but appears to have an extension method by that name.

Can also be called as

@Html.Partial("_PartialView", (ModelClass)View.Data)

Partial views in ASP.NET MVC – Rachel Appel, In ASP.NET MVC, a partial view is a custom, reusable component If you are using Razor as the default view engine, a partial view is simply In this case, your decision in choosing between RenderPartial or In this case, you want to give the partial view its own view model Michael Dann • 3 years ago. The Partial Views in asp.net mvc razor is similar to user control in ASP.NET Webforms. In asp.net mvc web application in case if we want to display some similar part of content in various part of web application then we need create a Partial View for that part.

There's another reason that this can be thrown, even if you're not using dynamic/ExpandoObject. If you are doing a loop, like this:

@foreach (var folder in ViewBag.RootFolder.ChildFolders.ToList())
{
    @Html.Partial("ContentFolderTreeViewItems", folder)
}

In that case, the "var" instead of the type declaration will throw the same error, despite the fact that RootFolder is of type "Folder. By changing the var to the actual type, the problem goes away.

@foreach (ContentFolder folder in ViewBag.RootFolder.ChildFolders.ToList())
{
    @Html.Partial("ContentFolderTreeViewItems", folder)
}

NET MVC and the Razor Engine that is used to render ASP. You can also pass a flag whether to render a partial view, which is quite common The context is the key that ties together the Razor View and controller, model, and the ASP. Listing 3 shows an example of using an MVC view to render error information from� ASP.NET MVC started off with the premise of being a very ‘pluggable’ framework and we can see it almost everywhere. You can plug in any IoC container, Testing Framework even View Engines! There are two view engines provided by default - the WebForms View engine (views with .aspx extention) and the Razor view engine (view with .cshtml

Here's a way to pass a dynamic object to a view (or partial view)

Add the following class anywhere in your solution (use System namespace, so its ready to use without having to add any references) -

    namespace System
    {
        public static class ExpandoHelper
        {
            public static ExpandoObject ToExpando(this object anonymousObject)
            {
                IDictionary<string, object> anonymousDictionary = HtmlHelper.AnonymousObjectToHtmlAttributes(anonymousObject);
                IDictionary<string, object> expando = new ExpandoObject();
                foreach (var item in anonymousDictionary)
                    expando.Add(item);
                return (ExpandoObject)expando;
            }

        }
    }

When you send the model to the view, convert it to Expando :

    return View(new {x=4, y=6}.ToExpando());

Cheers

NET MVC model, they'll naturally want to create reusable NET MVC 3 allow the developer to create reusable content with the very precise and clean Razor syntax (or ASPX). The syntax that renders the partial view is implemented as an Html helper. The Html.Partial helper renders the partial view named� This article explains how you can render partial view in main view with model data. You can render Partial View using Html.RenderAction and Html.RenderPartial helper methods. Create a new ASP.NET MVC application with name RenderPartialViewWithModel or other name which you like.

Partial Pages or Views are Razor files containing snippets of HTML and server-side code to be included in any number of pages or layouts. Partial pages can be used to break up complex pages into smaller units, thereby reducing the complexity and allowing teams to work on different units concurrently.

What Are Areas in ASP.Net MVC - Part 6; What is a Partial Views in ASP.NET MVC Partial view is just like a WEB User Control in ASP.NET web form technology. Partial views are used to componentize Razor views and make them easier to build and update. Partial views can also be returned directly from controller methods. In this case, the browser

Comments
  • Right, the main reason this doesn't work is that C# does not support calling an extension method (which is what Html.RenderPartial() is) when any of the arguments is of a dynamic type. You have to either call the extension method statically or cast the argument to a non-dynamic type.
  • +1 as seems more sensible to me than Diego's suggestion - for reasons noted above. i.e. if you know what type you are dealing with, then deal with that type !
  • doesn't make sense. If I want to use a dynamic model I would like that the helpers help me. Dynamic model in most cases is simpler and more productive since you don't have to declare the classes.
  • @ema - using dynamic models also leads to sloppier, poorly thought-out code. ViewModels are almost always a better idea than dynamic models. Unless you like finding compile errors at run-time!
  • @JoshM. This also happens when using ViewBag (aka dynamic model) which is suitable for passing around "client-side" variables, one-time use, etc. that really don't require the use or maintenance of a whole new ViewModel class.
  • @MattBorja Indeed. Dynamic stuff leaves room for error but can be really useful when used sparingly.
  • This has the downside that it generates a temporary (and potentially large) MvcHtmlString on the fly, rather than just writing to the output directly.
  • I found that I needed to cast my model like this, even though my model wasn't declared as a dynamic. It was probably because my model was a list.