using System;

using System.Collections;

using System.Collections.Generic;

using System.Linq;

using System.Text;

​

using Newtonsoft.Json;

using Newtonsoft.Json.Linq;

using Newtonsoft.Json.Converters;

using Newtonsoft.Json.Serialization;    

​

using NUnit.Framework; // Throws NUnit.Framework.AssertionException

​

//https://stackoverflow.com/questions/76291159/how-to-make-json-output-key-with-pascal-case-for-a-single-action-method?noredirect=1

​

    [JsonObject(NamingStrategyType = typeof(DefaultNamingStrategy))]

    public class AuthResponse

    {

        [JsonProperty("AuthenticationResult")]

        public AuthenticationResult authenticationResult { get; set; }

    }

​

    [JsonObject(NamingStrategyType = typeof(DefaultNamingStrategy))]

    public class AuthenticationResult

    {

        public string AccessToken { get; set; }

        public int ExpiresIn { get; set; }

        public string TokenType { get; set; }

        public string RefreshToken { get; set; }

        public string IdToken { get; set; }

    }

​

    class TestClass

    {

        public static void Test()

        {

            var AuthResponse = new AuthResponse 

            {

                authenticationResult = new() { AccessToken = "foo", ExpiresIn = 1, TokenType = "foo", RefreshToken = "foo", IdToken = "foo" },

            };

​

            var settings = new JsonSerializerSettings

            {

                ContractResolver = new CamelCasePropertyNamesContractResolver(),

            };

            var json = JsonConvert.SerializeObject(AuthResponse, Formatting.Indented, settings);

            Console.WriteLine(json);

        }   

    }

​

​

    public class Program

    {

        public static void Main()

        {

            Console.WriteLine("Environment version: {0} ({1}), {2}", System.Runtime.InteropServices.RuntimeInformation.FrameworkDescription , Environment.Version, Environment.OSVersion);

            Console.WriteLine("{0} version: {1}", typeof(JsonSerializer).Namespace, typeof(JsonSerializer).Assembly.FullName);

            Console.WriteLine();

​

            try

            {

                TestClass.Test();

            }

            catch (Exception ex)

            {

                Console.WriteLine("Failed with unhandled exception: ");

                Console.WriteLine(ex);

                throw;

            }

        }

    }

​