C# unsupported grant type when calling web api

Question

I am trying to perform a Post to my WebAPI from a c# WPF desktop app.

No matter what I do, I get

{"error":"unsupported_grant_type"}

This is what I've tried (and I've tried everything I could find):

Also dev web api currently active for testing: http://studiodev.biz/

base http client object:

var client = new HttpClient()
client.BaseAddress = new Uri("http://studiodev.biz/");
client.DefaultRequestHeaders.Accept.Clear();
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("text/plain"));
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("*/*"));

with the following send methods:

var response = await client.PostAsJsonAsync("token", "{'grant_type'='password'&'username'='username'&'password'='password'");
var response = await client.PostAsJsonAsync("token", "grant_type=password&username=username&password=password");

After that failed, I did some googling and tried:

LoginModel data = new LoginModel(username, password);
string json = JsonConvert.SerializeObject(data);
await client.PostAsync("token", new JsonContent(json));

same result, so I tried:

req.Content = new StringContent(json, Encoding.UTF8, "application/x-www-form-urlencoded");
await client.SendAsync(req).ContinueWith(respTask =>
{
 Application.Current.Dispatcher.Invoke(new Action(() => { label.Content = respTask.Result.ToString(); }));
});

Note: I can make a successful call with Chrome.

Update Fiddler Result

Could someone please help me make a successful call to the above web api... Please let me know if I can help clarify. Thanks!!

Answer 1

Here is a working example I used to make this request of my local Web API application running on port 43305 using SSL. I put the project on GitHub as well. https://github.com/casmer/WebAPI-getauthtoken

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using Newtonsoft.Json;
using System.Net.Http;
using System.Web;

namespace GetAccessTokenSample
{
  class Program
  {
    private static string baseUrl = "https://localhost:44305";

    static void Main(string[] args)
    {

      Console.WriteLine("Enter Username: ");
      string username= Console.ReadLine();
      Console.WriteLine("Enter Password: ");
      string password = Console.ReadLine();

      LoginTokenResult accessToken = GetLoginToken(username,password);
      if (accessToken.AccessToken != null)
      {
        Console.WriteLine(accessToken);
      }
      else
      {
        Console.WriteLine("Error Occurred:{0}, {1}", accessToken.Error, accessToken.ErrorDescription);
      }

    }


    private static LoginTokenResult GetLoginToken(string username, string password)
    {

      HttpClient client = new HttpClient();
      client.BaseAddress = new Uri(baseUrl);
      //TokenRequestViewModel tokenRequest = new TokenRequestViewModel() { 
      //password=userInfo.Password, username=userInfo.UserName};
      HttpResponseMessage response =
        client.PostAsync("Token",
          new StringContent(string.Format("grant_type=password&username={0}&password={1}",
            HttpUtility.UrlEncode(username),
            HttpUtility.UrlEncode(password)), Encoding.UTF8,
            "application/x-www-form-urlencoded")).Result;

      string resultJSON = response.Content.ReadAsStringAsync().Result;
      LoginTokenResult result = JsonConvert.DeserializeObject<LoginTokenResult>(resultJSON);

      return result;
    }

    public class LoginTokenResult
    {
      public override string ToString()
      {
        return AccessToken;
      }

      [JsonProperty(PropertyName = "access_token")]
      public string AccessToken { get; set; }

      [JsonProperty(PropertyName = "error")]
      public string Error { get; set; }

      [JsonProperty(PropertyName = "error_description")]
      public string ErrorDescription { get; set; }

    }

  }
}

Answer 2

If you are using RestSharp, you need to make the request like this:

public static U PostLogin<U>(string url, Authentication obj)
            where U : new()
        {
            RestClient client = new RestClient();
            client.BaseUrl = new Uri(host + url);
            var request = new RestRequest(Method.POST);
            string encodedBody = string.Format("grant_type=password&username={0}&password={1}",
                obj.username,obj.password);
            request.AddParameter("application/x-www-form-urlencoded", encodedBody, ParameterType.RequestBody);
            request.AddParameter("Content-Type", "application/x-www-form-urlencoded", ParameterType.HttpHeader);
            var response = client.Execute<U>(request);

             return response.Data;

        }

Answer 3

1) Note the URL: "localhost:55828/token" (not "localhost:55828/API/token")

2) Note the request data. Its not in json format, its just plain data without double quotes. "userName=[e-mail]&password=Test123$&grant_type=password"

3) Note the content type. Content-Type: 'application/x-www-form-urlencoded' (not Content-Type: 'application/json')

4) When you use javascript to make post request, you may use following:

$http.post("localhost:55828/token", 
    "userName=" + encodeURIComponent(email) +
        "&password=" + encodeURIComponent(password) +
        "&grant_type=password",
    {headers: { 'Content-Type': 'application/x-www-form-urlencoded' }}
).success(function (data) {//...

See screenshots below from Postman:

Answer 4

The default implementation of OAuthAuthorizationServerHandler only accepts form encoding (i.e. application/x-www-form-urlencoded) and not JSON encoding (application/JSON).

Your request's ContentType should be application/x-www-form-urlencoded and pass the data in the body as:

grant_type=password&username=Alice&password=password123

i.e. not in JSON format.

The chrome example above works because it is not passing data as JSON. You only need this for getting a token; for other methods of your API you can use JSON.

This kind of problem is also discussed here.