Asked  7 Months ago    Answers:  5   Viewed   116 times

I'm working on an existing application that has been partially converted over to MVC. Whenever a controller responds with a JSON ActionResult, the enums are sent as numbers opposed to the string name. It sounds like the default serializer should be JSON.Net, which should be sending the enums over as their names opposed to the integer representation, but that's not the case here.

Am I missing a web.config setting that sets this as the default serializer? Or is there another setting that needs to be changed?



In ASP.Net MVC4 the default JavaScript serializer which is used in the JsonResult class is still the JavaScriptSerializer (you can check it in the code)

I think you have confused it with the ASP.Net Web.API where JSON.Net is the default JS serializer but MVC4 doesn't use it.

So you need to configure JSON.Net to work with MVC4 (basically you need to create your own JsonNetResult), there are plenty of articles about it:

  • ASP.NET MVC and Json.NET
  • Using JSON.NET as the default JSON serializer in ASP.NET MVC 3 - is it possible?

If you also want to use JSON.Net for controller action parameters so during the model binding then you need write your own ValueProviderFactory implementation.

And you need to register your implementation with:

ValueProviderFactories.Factories.Add(new MyJsonValueProviderFactory());

You can use the built in JsonValueProviderFactory as an example or this article: ASP.NET MVC 3 – Improved JsonValueProviderFactory using Json.Net

Tuesday, June 1, 2021
answered 7 Months ago

Your model is null because the way you're supplying the inputs to your form means the model binder has no way to distinguish between the elements. Right now, this code:

@foreach (var planVM in Model)
    @Html.Partial("_partialView", planVM)

is not supplying any kind of index to those items. So it would repeatedly generate HTML output like this:

<input type="hidden" name="yourmodelprefix.PlanID" />
<input type="hidden" name="yourmodelprefix.CurrentPlan" />
<input type="checkbox" name="yourmodelprefix.ShouldCompare" />

However, as you're wanting to bind to a collection, you need your form elements to be named with an index, such as:

<input type="hidden" name="yourmodelprefix[0].PlanID" />
<input type="hidden" name="yourmodelprefix[0].CurrentPlan" />
<input type="checkbox" name="yourmodelprefix[0].ShouldCompare" />
<input type="hidden" name="yourmodelprefix[1].PlanID" />
<input type="hidden" name="yourmodelprefix[1].CurrentPlan" />
<input type="checkbox" name="yourmodelprefix[1].ShouldCompare" />

That index is what enables the model binder to associate the separate pieces of data, allowing it to construct the correct model. So here's what I'd suggest you do to fix it. Rather than looping over your collection, using a partial view, leverage the power of templates instead. Here's the steps you'd need to follow:

  1. Create an EditorTemplates folder inside your view's current folder (e.g. if your view is HomeIndex.cshtml, create the folder HomeEditorTemplates).
  2. Create a strongly-typed view in that directory with the name that matches your model. In your case that would be PlanCompareViewModel.cshtml.

Now, everything you have in your partial view wants to go in that template:

@model PlanCompareViewModel
    @Html.HiddenFor(p => p.PlanID)
    @Html.HiddenFor(p => p.CurrentPlan)
    @Html.CheckBoxFor(p => p.ShouldCompare)
   <input type="submit" value="Compare"/>

Finally, your parent view is simplified to this:

@model IEnumerable<PlanCompareViewModel>
@using (Html.BeginForm("ComparePlans", "Plans", FormMethod.Post, new { id = "compareForm" }))

DisplayTemplates and EditorTemplates are smart enough to know when they are handling collections. That means they will automatically generate the correct names, including indices, for your form elements so that you can correctly model bind to a collection.

Tuesday, June 1, 2021
answered 7 Months ago

I believe the best way to do it, is - as described in your links - to extend ActionResult or extend JsonResult directly.

As for the method JsonResult that is not virtual on the controller that's not true, just choose the right overload. This works well:

protected override JsonResult Json(object data, string contentType, Encoding contentEncoding)

EDIT 1: A JsonResult extension...

public class JsonNetResult : JsonResult
    public override void ExecuteResult(ControllerContext context)
        if (context == null)
            throw new ArgumentNullException("context");

        var response = context.HttpContext.Response;

        response.ContentType = !String.IsNullOrEmpty(ContentType) 
            ? ContentType 
            : "application/json";

        if (ContentEncoding != null)
            response.ContentEncoding = ContentEncoding;

        // If you need special handling, you can call another form of SerializeObject below
        var serializedObject = JsonConvert.SerializeObject(Data, Formatting.Indented);

EDIT 2: I removed the check for Data being null as per the suggestions below. That should make newer versions of JQuery happy and seems like the sane thing to do, as the response can then be unconditionally deserialized. Be aware though, that this is not the default behavior for JSON responses from ASP.NET MVC, which rather responds with an empty string, when there's no data.

Tuesday, June 1, 2021
answered 7 Months ago

There's an overload for the SelectList constructor that takes 4 arguments. The last of which is the default selected object. E.g:

ViewBag.Vessels = new SelectList(vs, "InstId", "InstName", selectedValue);

Where selectedValue is an object of whatever type is in your list.

Sunday, August 15, 2021
answered 4 Months ago

WebApi is an alternative Service oriented application from Microsoft just like WCF. But WCF uses SOAP protocol and WebAPI uses HTTP protocol for communication.

So if you are using WCF to provide service for your MVC application you would host that wcf service seperately and consume its service by MVC application, EXACTLY same way you have to host your WebAPI project seperately and provide service to your Web application (MVC).

for some reasons if you want them (MVC and WebAPI) to use in the same project, follow this rules from this article.

Thursday, September 9, 2021
answered 3 Months ago
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