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 = JsonConvert.Deserialize(json);
Console.WriteLine(jsonResponse.message);
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 = JsonConvert.Deserialize(json);
Console.WriteLine(jsonResponse.message);
Json.NET allows us to do this:
dynamic d = JObject.Parse("{number:1000, str:'string', array: [1,2,3,4,5,6]}");
Console.WriteLine(d.number);
Console.WriteLine(d.str);
Console.WriteLine(d.array.Count);
Output:
1000
string
6
Documentation here: LINQ to JSON with Json.NET
See also JObject.Parse and JArray.Parse
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.
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.
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:
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.
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>();