Deserialize JSON with C#

Question

I'm trying to deserialize a Facebook friend's Graph API call into a list of objects. The JSON object looks like:

{"data":[{"id":"518523721","name":"ftyft"},
         {"id":"527032438","name":"ftyftyf"},
         {"id":"527572047","name":"ftgft"},
         {"id":"531141884","name":"ftftft"},
         {"id":"532652067","name"...

List<EFacebook> facebooks = new JavaScriptSerializer().Deserialize<List<EFacebook>>(result);

It's not working, because the primitive object is invalid. How can I deserialize this?

Answer 1

You need to create a structure like this:

public class Friends
{

    public List<FacebookFriend> data {get; set;}
}

public class FacebookFriend
{

    public string id {get; set;}
    public string name {get; set;}
}

Then you should be able to do:

Friends facebookFriends = new JavaScriptSerializer().Deserialize<Friends>(result);

The names of my classes are just an example. You should use proper names.

Adding a sample test:

string json =
    @"{""data"":[{""id"":""518523721"",""name"":""ftyft""}, {""id"":""527032438"",""name"":""ftyftyf""}, {""id"":""527572047"",""name"":""ftgft""}, {""id"":""531141884"",""name"":""ftftft""}]}";

Friends facebookFriends = new System.Web.Script.Serialization.JavaScriptSerializer().Deserialize<Friends>(json);

foreach(var item in facebookFriends.data)
{
    Console.WriteLine("id: {0}, name: {1}", item.id, item.name);
}

Produces:

id: 518523721, name: ftyft
id: 527032438, name: ftyftyf
id: 527572047, name: ftgft
id: 531141884, name: ftftft

Answer 2

Sometimes I prefer dynamic objects:

public JsonResult GetJson()
{
  string res;
  WebClient client = new WebClient();

  // Download string
  string value = client.DownloadString("https://api.instagram.com/v1/users/000000000/media/recent/?client_id=clientId");

  // Write values
  res = value;
  dynamic dyn = JsonConvert.DeserializeObject(res);
  var lstInstagramObjects = new List<InstagramModel>();

  foreach(var obj in dyn.data)
  {
    lstInstagramObjects.Add(new InstagramModel()
    {
      Link = (obj.link != null) ? obj.link.ToString() : "",
      VideoUrl = (obj.videos != null) ? obj.videos.standard_resolution.url.ToString() : "",
      CommentsCount = int.Parse(obj.comments.count.ToString()),
      LikesCount = int.Parse(obj.likes.count.ToString()),
      CreatedTime = new System.DateTime(1970, 1, 1, 0, 0, 0, 0).AddSeconds((double.Parse(obj.created_time.ToString()))),
      ImageUrl = (obj.images != null) ? obj.images.standard_resolution.url.ToString() : "",
      User = new InstagramModel.UserAccount()
             {
               username = obj.user.username,
               website = obj.user.website,
               profile_picture = obj.user.profile_picture,
               full_name = obj.user.full_name,
               bio = obj.user.bio,
               id = obj.user.id
             }
    });
  }

  return Json(lstInstagramObjects, JsonRequestBehavior.AllowGet);
}

Answer 3

A great way to automatically generate these classes for you is to copy your JSON output and throw it in here:

http://json2csharp.com/

It will provide you with a starting point to touch up your classes for deserialization.

Answer 4

Very easily we can parse JSON content with the help of dictionary and JavaScriptSerializer. Here is the sample code by which I parse JSON content from an ashx file.

var jss = new JavaScriptSerializer();
string json = new StreamReader(context.Request.InputStream).ReadToEnd();
Dictionary<string, string> sData = jss.Deserialize<Dictionary<string, string>>(json);
string _Name = sData["Name"].ToString();
string _Subject = sData["Subject"].ToString();
string _Email = sData["Email"].ToString();
string _Details = sData["Details"].ToString();

Answer 5

Newtonsoft.JSON is a good solution for these kind of situations. Also Newtonsof.JSON is faster than others, such as JavaScriptSerializer, DataContractJsonSerializer.

In this sample, you can the following:

var jsonData = JObject.Parse("your JSON data here");

Then you can cast jsonData to JArray, and you can use a for loop to get data at each iteration.

Also, I want to add something:

for (int i = 0; (JArray)jsonData["data"].Count; i++)
{
    var data = jsonData[i - 1];
}

Working with dynamic object and using Newtonsoft serialize is a good choice.

Answer 6

I agree with Icarus (would have commented if I could), but instead of using a CustomObject class, I would use a Dictionary (in case Facebook adds something).

private class MyFacebookClass
{
    public IList<IDictionary<string, string>> data { get; set; }
}

or

private class MyFacebookClass
{
    public IList<IDictionary<string, object>> data { get; set; }
}

Answer 7

Serialization:

// Convert an object to JSON string format
string jsonData = JsonConvert.SerializeObject(obj);

Response.Write(jsonData);

Deserialization::

To deserialize a dynamic object

  string json = @"{
      'Name': 'name',
      'Description': 'des'
    }";

var res = JsonConvert.DeserializeObject< dynamic>(json);

Response.Write(res.Name);

Answer 8

If you're using .NET Core 3.0, you can use System.Text.Json (which is now built-in) to deserialize JSON.

The first step is to create classes to model the JSON. There are many tools which can help with this, and some of the answers here list them.

Some options are http://json2csharp.com, http://app.quicktype.io, or use Visual Studio (menu Edit → Paste Special → Paste JSON as classes).

public class Person
{
    public string Id { get; set; }
    public string Name { get; set; }
}

public class Response
{
    public List<Person> Data { get; set; }
}

Then you can deserialize using:

var people = JsonSerializer.Deserialize<Response>(json);

---

If you need to add settings, such as camelCase handling, then pass serializer settings into the deserializer like this:

var options = new JsonSerializerOptions() { PropertyNamingPolicy = JsonNamingPolicy.CamelCase };
var person = JsonSerializer.Deserialize<Response>(json, options);

Answer 9

You can use this extensions

public static class JsonExtensions
{
   public static T ToObject<T>(this string jsonText)
   {
       return JsonConvert.DeserializeObject<T>(jsonText);
   }

   public static string ToJson<T>(this T obj)
   {
       return JsonConvert.SerializeObject(obj);
   }
}

Answer 10

Here is another site that will help you with all the code you need as long as you have a correctly formated JSON string available:

https://app.quicktype.io/