In ASP.NET MVC, deserialize JSON prior to or in controller's action method

Question

I am working on a website that will post a JSON object (using jQuery Post method) to the server side.

{ 
    "ID" : 1,
    "FullName" : {
       "FirstName" : "John",
       "LastName" : "Smith"
    }
}

At the same time, I wrote classes on the server side for this data structure.

public class User
{
    public int ID { get; set; }
    public Name FullName { get; set;}
}

public class Name
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

When I run the website with following code in my controller class, the FullName property doesn't get deserialized. What am I doing wrong?

[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Submit(User user)
{
    // At this point, user.FullName is NULL. 

    return View();
}

Answer 1

I resolved my problem by implementing an action filter; code sample is provided below. From the research, I learned that there is another solution, model binder, as takepara described above. But I don't really know that pros and cons of doing in either approach.

Thanks to Steve Gentile's blog post for this solution.

public class JsonFilter : ActionFilterAttribute
    {
        public string Parameter { get; set; }
        public Type JsonDataType { get; set; }

        public override void OnActionExecuting(ActionExecutingContext filterContext)
        {
            if (filterContext.HttpContext.Request.ContentType.Contains("application/json"))
            {
                string inputContent;
                using (var sr = new StreamReader(filterContext.HttpContext.Request.InputStream))
                {
                    inputContent = sr.ReadToEnd();
                }

                var result = JsonConvert.DeserializeObject(inputContent, JsonDataType);
                filterContext.ActionParameters[Parameter] = result;
            }
        }
    }

[AcceptVerbs(HttpVerbs.Post)]
[JsonFilter(Parameter="user", JsonDataType=typeof(User))]
public ActionResult Submit(User user)
{
    // user object is deserialized properly prior to execution of Submit() function

    return View();
}