Dynamic variable not working in C# with Json.Net

Question

Can anyone tell me why I get an error when trying to output"dJson2.Type" in the code below?

    string Json1= @"[{'Id':1, 'FirstName':'John', 'LastName':'Smith'}, {'Id':2, 'FirstName':'Jane', 'LastName':'Doe'}]";
    dynamic dJson1= JsonConvert.DeserializeObject(Json1);
    Console.WriteLine(dJson1.GetType());
    Console.WriteLine(dJson1.Type);

    string Json2 = @"{'Id':1, 'FirstName':'John', 'LastName':'Smith'}";
    dynamic dJson2 = JsonConvert.DeserializeObject(Json2);
    Console.WriteLine(dJson2.GetType());
    Console.WriteLine(dJson2.Type);

The program dies on the Console.WriteLine(dJson2.Type) statement. The output of the program is...

Newtonsoft.Json.Linq.JArray
Array
Newtonsoft.Json.Linq.JObject
(should say Object here, I think)

Inspecting the local variables, dJson2 has a "Type" property with value "Object".

Yeah, the error message should tell you exactly what's wrong. — Robert Harvey
– Robert Harvey, Commented Sep 30, 2016 at 22:29
The call is ambiguous between the following methods or properties: 'System.Console.WriteLine(string, params object[])' — Drew Kennedy
– Drew Kennedy, Commented Sep 30, 2016 at 22:30
I'm taking a shot in the dark here - I think in the first example the compiler is smart enough to figure out it's an array. But in the second example it's looking for a property called Type, and it doesn't exist. Borrowing your code, as soon as I explicitly put in a property called Type in your JSON string, it compiled as expected. — Drew Kennedy
– Drew Kennedy, Commented Sep 30, 2016 at 22:37
In answer to Stanley/Harvey. The error actually doesn't help much. The problem is that the Type property if dJson2 is coming back with a null value. You can, for instance, add "var jType = dJson2.Type" in the code and you will find that jType is null. — Boulder Keith
– Boulder Keith, Commented Oct 1, 2016 at 1:10

DolphinSX · Accepted Answer · 2016-10-01 14:47:11Z

1

This is because JObject behaves similarly as System.Dynamic.ExpandoObject. Try to change your example to:

  string Json2 = @"{'Id':1, 'FirstName':'John', 'LastName':'Smith'}";
  dynamic dJson2 = JsonConvert.DeserializeObject(Json2);
  dJson2.Type = "mynewfield";
  Console.WriteLine(dJson2.GetType());
  Console.WriteLine(dJson2.Type);

If you want to get property of underlying type you need to cast it (to JToken or JObject), otherwise requested property will be searched in IDictionary<string, JToken> that JObject implements.

This example may help:

  dynamic oobj = new JObject();
  oobj.Type = "TEST";
  Console.WriteLine(oobj.Type);
  Console.WriteLine(((JObject)oobj).Type);

answered Oct 1, 2016 at 14:47

DolphinSX

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1 Comment

Boulder Keith Over a year ago

Huh. Makes sense, though I'm not sure the behavior is what I like or what I would expect. It's weird that it works for JArray but not JObject. Anyway, a cast definitely works. What I ended out doing was a switch on dJson2.GetType(), casting to the actual type, then moving on. This behaves as expected, if somewhat verbose.... Thanks for explaining, DolphinSX.

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