Question

I have the following arbitrary JSON object(The field names may be changed).

  {
    firstname: "Ted",
    lastname: "Smith",
    age: 34,
    married : true
  }

–

public JsonResult GetData(??????????){
.
.
.
}

I know that I can define a class just like the JSON object with the same field names as the argument, But I would like my controller to accept arbitrary JSON object with different field names.

Answer 1

If you want to pass custom JSON object to MVC action then you can use this solution, it works like a charm.

    public string GetData()
    {
        // InputStream contains the JSON object you've sent
        String jsonString = new StreamReader(this.Request.InputStream).ReadToEnd();

        // Deserialize it to a dictionary
        var dic = 
          Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<String, dynamic>>(jsonString);

        string result = "";

        result += dic["firstname"] + dic["lastname"];

        // You can even cast your object to their original type because of 'dynamic' keyword
        result += ", Age: " + (int)dic["age"];

        if ((bool)dic["married"])
            result += ", Married";


        return result;
    }

The real benefit of this solution is that you don’t require to define a new class for each combination of arguments and beside that, you can cast your objects to their original types easily.

UPDATED

Now, you can even merge your GET and POST action methods since your post method doesn’t have any argument any more just like this :

 public ActionResult GetData()
 {
    // GET method
    if (Request.HttpMethod.ToString().Equals("GET"))
        return View();

    // POST method 
    .
    .
    .

    var dic = GetDic(Request);
    .
    .
    String result = dic["fname"];

    return Content(result);
 }

and you can use a helper method like this to facilitate your job

public static Dictionary<string, dynamic> GetDic(HttpRequestBase request)
{
    String jsonString = new StreamReader(request.InputStream).ReadToEnd();
    return Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<string, dynamic>>(jsonString);
}