Deserialize json object into dynamic object using Json.net

Question

Is it possible to return a dynamic object from a json deserialization using json.net? I would like to do something like this:

dynamic jsonResponse = JsonConv         


        

Answer 1

Note: At the time I answered this question in 2010, there was no way to deserialize without some sort of type, this allowed you to deserialize without having go define the actual class and allowed an anonymous class to be used to do the deserialization.

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You need to have some sort of type to deserialize to. You could do something along the lines of:

var product = new { Name = "", Price = 0 };
dynamic jsonResponse = JsonConvert.Deserialize(json, product.GetType());

My answer is based on a solution for .NET 4.0's build in JSON serializer. Link to deserialize to anonymous types is here:

http://blogs.msdn.com/b/alexghi/archive/2008/12/22/using-anonymous-types-to-deserialize-json-data.aspx

Answer 2

As of Json.NET 4.0 Release 1, there is native dynamic support:

[Test]
public void DynamicDeserialization()
{
    dynamic jsonResponse = JsonConvert.DeserializeObject("{\"message\":\"Hi\"}");
    jsonResponse.Works = true;
    Console.WriteLine(jsonResponse.message); // Hi
    Console.WriteLine(jsonResponse.Works); // True
    Console.WriteLine(JsonConvert.SerializeObject(jsonResponse)); // {"message":"Hi","Works":true}
    Assert.That(jsonResponse, Is.InstanceOf<dynamic>());
    Assert.That(jsonResponse, Is.TypeOf<JObject>());
}

And, of course, the best way to get the current version is via NuGet.

Updated (11/12/2014) to address comments:

This works perfectly fine. If you inspect the type in the debugger you will see that the value is, in fact, dynamic. The underlying type is a JObject. If you want to control the type (like specifying ExpandoObject, then do so.

Answer 3

I know this is old post but JsonConvert actually has a different method so it would be

var product = new { Name = "", Price = 0 };
var jsonResponse = JsonConvert.DeserializeAnonymousType(json, product);

Answer 4

If you use JSON.NET with old version which didn't JObject.

This is another simple way to make a dynamic object from JSON: https://github.com/chsword/jdynamic

NuGet Install

PM> Install-Package JDynamic

Support using string index to access member like:

dynamic json = new JDynamic("{a:{a:1}}");
Assert.AreEqual(1, json["a"]["a"]);

Test Case

And you can use this util as following :

Get the value directly

dynamic json = new JDynamic("1");

//json.Value

2.Get the member in the json object

dynamic json = new JDynamic("{a:'abc'}");
//json.a is a string "abc"

dynamic json = new JDynamic("{a:3.1416}");
//json.a is 3.1416m

dynamic json = new JDynamic("{a:1}");
//json.a is integer: 1

3.IEnumerable

dynamic json = new JDynamic("[1,2,3]");
/json.Length/json.Count is 3
//And you can use json[0]/ json[2] to get the elements

dynamic json = new JDynamic("{a:[1,2,3]}");
//json.a.Length /json.a.Count is 3.
//And you can use  json.a[0]/ json.a[2] to get the elements

dynamic json = new JDynamic("[{b:1},{c:1}]");
//json.Length/json.Count is 2.
//And you can use the  json[0].b/json[1].c to get the num.

Other

dynamic json = new JDynamic("{a:{a:1} }");

//json.a.a is 1.

Answer 5

Yes it is possible. I have been doing that all the while.

dynamic Obj = JsonConvert.DeserializeObject(<your json string>);

It is a bit trickier for non native type. Suppose inside your Obj, there is a ClassA, and ClassB objects. They are all converted to JObject. What you need to do is:

ClassA ObjA = Obj.ObjA.ToObject<ClassA>();
ClassB ObjB = Obj.ObjB.ToObject<ClassB>();

Answer 6

Yes you can do it using the JsonConvert.DeserializeObject. To do that, just simple do:

dynamic jsonResponse = JsonConvert.DeserializeObject(json);
Console.WriteLine(jsonResponse["message"]);