Deserialize array of key value pairs using Json.NET

Question

Given the following json:

[ {\"id\":\"123\", ... \"data\":[{\"key1\":\"val1\"}, {\"key2\":\"val2\"}], ...}, ... ]

that is part of a bigger

Answer 1

Thanks @Boo for your answer but in my case I needed to take some small adjustements. This is how my JSON looks like:

{
    "rates": {
        "CAD": 1.5649,
        "CZK": 26,118,
        ...
    },
    "base": "EUR",
    "date": "2020-08-16"
}

And my DTO looks like the following:

public IDictionary<string, decimal> Rates { get; set; }
public string Base { get; set; }
public DateTime Date { get; set; }

So the only adjustement was to remove the IEnumerable around the IDictionary.

Answer 2

I ended up doing this:

[JsonConverter(typeof(MyCustomClassConverter))]
public class MyCustomClass
{
  internal class MyCustomClassConverter : JsonConverter
  {
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
      throw new NotImplementedException();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
      JObject jObject = JObject.Load(reader);

      foreach (var prop in jObject)
      {
        return new MyCustomClass { Key = prop.Key, Value = prop.Value.ToString() };
      }

      return null;
    }

    public override bool CanConvert(Type objectType)
    {
      return typeof(MyCustomClass).IsAssignableFrom(objectType);
    }
  }

  public string Key { get; set; }
  public string Value { get; set; }
}

Answer 3

public class Datum
{
    public string key1 { get; set; }
    public string key2 { get; set; }
}

public class RootObject
{
    public string id { get; set; }
    public List<Datum> data { get; set; }
}

i used this wizard as well json2csharp.com to generate class for deserialized

for using that

using RestSharp;
using Newtonsoft.Json;

IRestResponse restSharp= callRestGetMethodby_restSharp(api_server_url);
string jsonString= restSharp.Content;

RootObject rootObj= JsonConvert.DeserializeObject<RootObject>(jsonString);
return Json(rootObj);

if you call rest by restsharp

    public IRestResponse callRestGetMethodby_restSharp(string API_URL)
    {
        var client = new RestSharp.RestClient(API_URL);
        var request = new RestRequest(Method.GET);
        request.AddHeader("Content-Type", "application/json");
        request.AddHeader("cache-control", "no-cache");
        IRestResponse response = client.Execute(request);
        return response;
    }

also you can get this 6 line of restsharp from getpostman.com tools

Answer 4

The simplest way is deserialize array of key-value pairs to IDictionary<string, string>:

public class SomeData
{
    public string Id { get; set; }

    public IEnumerable<IDictionary<string, string>> Data { get; set; }
}

private static void Main(string[] args)
{
    var json = "{ \"id\": \"123\", \"data\": [ { \"key1\": \"val1\" }, { \"key2\" : \"val2\" } ] }";

    var obj = JsonConvert.DeserializeObject<SomeData>(json);
}

But if you need deserialize that to your own class, it can be looks like that:

public class SomeData2
{
    public string Id { get; set; }

    public List<SomeDataPair> Data { get; set; }
}

public class SomeDataPair
{
    public string Key { get; set; }

    public string Value { get; set; }
}

private static void Main(string[] args)
{
    var json = "{ \"id\": \"123\", \"data\": [ { \"key1\": \"val1\" }, { \"key2\" : \"val2\" } ] }";

    var rawObj = JObject.Parse(json);

    var obj2 = new SomeData2
    {
        Id = (string)rawObj["id"],
        Data = new List<SomeDataPair>()
    };

    foreach (var item in rawObj["data"])
    {
        foreach (var prop in item)
        {
            var property = prop as JProperty;

            if (property != null)
            {
                obj2.Data.Add(new SomeDataPair() { Key = property.Name, Value = property.Value.ToString() });
            }

        }
    }
}

See that I khow that Value is string and i call ToString() method, there can be another complex class.