Congratulations on delving into the world of MVC and uncovering one of its initial drawbacks.

The process of converting data to JSON in either the view or controller may seem illogical, but sometimes you have no choice but to work within these constraints.

A practical solution I've come across is sending the JSON to the view using a ViewModel, as shown below:

var data = somedata;
var viewModel = new ViewModel();
var serializer = new JavaScriptSerializer();
viewModel.JsonData = serializer.Serialize(data);

return View("viewname", viewModel);

Then, utilize <%= Model.JsonData %> in your view. Just keep in mind that the standard .NET JavaScriptSerializer is not the most reliable tool out there.

If you opt to handle this conversion in the controller, it does make testing easier (although with slight modifications - consider incorporating an ISerializer dependency for mocking).

Update: In addition, when dealing with your JavaScript code, it's advisable to encapsulate all widget JS within parentheses like so:

(
    // all js here
)();

This practice helps prevent conflicts if you have multiple widgets on a single page. While it may not be necessary now, adopting this approach early can save you significant effort in the future. Furthermore, it aligns with object-oriented principles by encapsulating functionality effectively.

Implementing it this way in the view felt quite satisfying:

    @Html.HiddenJsonFor(m => m.TrackingTypes)

Below is the corresponding helper method Extension class:

public static class DataHelpers
{
    public static MvcHtmlString HiddenJsonFor<TModel, TProperty>(this HtmlHelper<TModel> htmlHelper, Expression<Func<TModel, TProperty>> expression)
    {
        return HiddenJsonFor(htmlHelper, expression, (IDictionary<string, object>) null);
    }

    public static MvcHtmlString HiddenJsonFor<TModel, TProperty>(this HtmlHelper<TModel> htmlHelper, Expression<Func<TModel, TProperty>> expression, object htmlAttributes)
    {
        return HiddenJsonFor(htmlHelper, expression, HtmlHelper.AnonymousObjectToHtmlAttributes(htmlAttributes));
    }

    public static MvcHtmlString HiddenJsonFor<TModel, TProperty>(this HtmlHelper<TModel> htmlHelper, Expression<Func<TModel, TProperty>> expression, IDictionary<string, object> htmlAttributes)
    {
        var name = ExpressionHelper.GetExpressionText(expression);
        var metadata = ModelMetadata.FromLambdaExpression(expression, htmlHelper.ViewData);

        var tagBuilder = new TagBuilder("input");
        tagBuilder.MergeAttributes(htmlAttributes);
        tagBuilder.MergeAttribute("name", name);
        tagBuilder.MergeAttribute("type", "hidden");

        var json = JsonConvert.SerializeObject(metadata.Model);

        tagBuilder.MergeAttribute("value", json);

        return MvcHtmlString.Create(tagBuilder.ToString());
    }
}

It may not be the most advanced solution, but it effectively resolves the dilemma of where to place it (in Controller or in view?) The answer: neither ;)

To access Json directly from the action, you can follow this example:

Your action method may look something like this:

virtual public JsonResult DisplaySomeWidget(int id)
{
    SomeModelView returnData = someDataMapper.getbyid(id);
    return Json(returnData);
}

Note

If you are assuming that this is the Model for a View, then the above code might not be entirely correct. You would need to make an Ajax call to the controller method in order to retrieve the data. In this case, the ascx would not have a model by itself. I will keep my code as it may still be helpful, and you can adjust the call accordingly.

Update 2 Remember to include the 'id' parameter in the code provided above.

To transform the View Modal Object into JSON format, utilize @Html.Raw(Json.Encode(object))

Expanding upon a fantastic response provided by Dave, it is possible to develop a simple HtmlHelper.

public static IHtmlString RenderAsJson(this HtmlHelper helper, object model)
{
    return helper.Raw(Json.Encode(model));
}

Implement this in your view:

@Html.RenderAsJson(Model)

This technique enables you to consolidate the process of generating JSON, facilitating any future modifications to the logic if necessary.

If you are working with ASP.NET Core 3.1 and above, there is a convenient way to serialize ViewModel to JSON using the System.Text.Json namespace:

@using System.Text.Json

...

@section Scripts {
    <script>
        const modelData = @Html.Raw(JsonSerializer.Serialize(Model));
    </script>
}

<htmltag id='elementId' data-AAAAA='@Html.Raw(Json.Encode(Model))' />

For more information, please visit this link

I followed the steps below and it worked perfectly.

<input id="hdnElement" class="hdnElement" type="hidden" value='@Html.Raw(Json.Encode(Model))'>

Although Andrew provided a great response, I wanted to make a few adjustments. My approach differs in that I prefer my ModelViews to contain only data related to the object, without any additional overhead information. It appears that using ViewData can add unnecessary data, but as a newcomer to this concept, I may be mistaken. Here is a suggestion:

Controller

virtual public ActionResult DisplaySomeWidget(int id)
{
    SomeModelView returnData = someDataMapper.getbyid(1);
    var serializer = new JavaScriptSerializer();
    ViewData["JSON"] = serializer.Serialize(returnData);
    return View(myview, returnData);
}

View

//create base js object;
var myWidget= new Widget(); //Widget is a class with a public member variable called data.
myWidget.data= <%= ViewData["JSON"] %>;

This setup ensures that the JSON data matches the data in your ModelView, allowing you to potentially send the JSON back to your controller with all components intact. While similar to making a request via JSONRequest, it eliminates one extra call, thereby reducing unnecessary overhead. However, handling dates might present challenges best discussed in another thread.