Deserialize JSON Object C# with Newtonsoft

Question

I need to deserialize the following:

{"result":{"success":true,"value":"8cb2237d0679ca88db6464eac60da96345513964"}}

to a C# object using Newtonsoft.Json

WebClient wc = new WebClient();
var json = wc.DownloadString(url);
Worker w = JsonConvert.DeserializeObject<Worker>(json);

Here is the class code:

public class Worker
{

    [JsonProperty("success")]
    public string success { get; set; }

    [JsonProperty("value")]
    public string value { get; set; }
}

The code does not error out, but the success and value are null.

luksan · Accepted Answer · 2013-01-28 20:10:11Z

5

You're missing the outer object.

public class Worker
{
     [JsonProperty("result")]
     public Result Result { get; set; }
}

public class Result
{
    [JsonProperty("success")]
    public string Success { get; set; }

    [JsonProperty("value")]
    public string Value { get; set; }
}

edited Jan 28, 2013 at 20:10

answered Jan 28, 2013 at 19:56

luksan

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6 Comments

user2019423 Over a year ago

Thanks! Figured "result" needed to be serialized.

luksan Over a year ago

Also you may want to consider capitalizing your property names, since that is the standard convention in C#. Since you're already controlling the JSON serialization with attributes, it won't affect that. I edited my answer to show what I mean.

anar khalilov Over a year ago

@luksan +1 What if I want the C# class to contain three fields: Result, Success and Value? When I set the JsonProperty attribute to "result.success" or "result.value" it returns null. Is it possible?

luksan Over a year ago

@Anar I'm not sure I understand. Maybe try creating a new question on here with example code.

anar khalilov Over a year ago

My fault, sorry. I wanted to ask whether you could specify JsonProperty value two levels deep, hence, "result.success". It does not work that way, maybe you know a way.

|

I4V · Accepted Answer · 2013-01-28 19:58:17Z

0

You don't need any class and can make use of dynamic keyword

string json = @"{""result"":{""success"":true,""value"":""8cb2237d0679ca88db6464eac60da96345513964""}}";

dynamic dynObj = JsonConvert.DeserializeObject(json);
Console.WriteLine("{0} {1}", dynObj.result.success, dynObj.result.value);

answered Jan 28, 2013 at 19:58

I4V

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Comments

MrLeap · Accepted Answer · 2013-01-28 20:14:39Z

0

I'm not familiar with that library, but success and result look to be both properties of the object "result"

Have you tried [JsonProperty("result.success")]?

Edit: Well, regardless it looks like a scoping issue. After viewing the documentation, this is my new suggestion:

public class Result{
 [JsonProperty("result")]
 public Worker result { get; set; }
}

then Json.Convert.Deserialize<Result>(json) instead.

edited Jan 28, 2013 at 20:14

answered Jan 28, 2013 at 19:56

MrLeap

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1 Comment

user2019423 Over a year ago

Good thought... still Null.

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