How to POST request using RestSharp

Question

I m trying to POST the request using RestSharp client as follows I m passing the Auth Code to following function

public void ExchangeCodeForToken(string code)
{
    if (string.IsNullOrEmpty(code))
    {
        OnAuthenticationFailed();
    }
    else
    {           
        var request = new RestRequest(this.TokenEndPoint, Method.POST);
        request.AddParameter("code", code);
        request.AddParameter("client_id", this.ClientId);
        request.AddParameter("client_secret", this.Secret);
        request.AddParameter("redirect_uri", "urn:ietf:wg:oauth:2.0:oob");
        request.AddParameter("grant_type", "authorization_code");
        request.AddHeader("content-type", "application/x-www-form-urlencoded");

        client.ExecuteAsync<AuthResult>(request, GetAccessToken);
    }
}

void GetAccessToken(IRestResponse<AuthResult> response)
{
    if (response == null || response.StatusCode != HttpStatusCode.OK
                         || response.Data == null 
                         || string.IsNullOrEmpty(response.Data.access_token))
    {
        OnAuthenticationFailed();
    }
    else
    {
        Debug.Assert(response.Data != null);
        AuthResult = response.Data;
        OnAuthenticated();
    }
}

But i am getting response.StatusCode = Bad Request. Can anyone help me for how do i POST the request using Restsharp client.

score 66 · Answer 1 · edited Mar 30 '16 at 14:12

66

My RestSharp POST method:

var client = new RestClient(ServiceUrl);

var request = new RestRequest("/resource/", Method.POST);

// Json to post.
string jsonToSend = JsonHelper.ToJson(json);

request.AddParameter("application/json; charset=utf-8", jsonToSend, ParameterType.RequestBody);
request.RequestFormat = DataFormat.Json;

try
{
    client.ExecuteAsync(request, response =>
    {
        if (response.StatusCode == HttpStatusCode.OK)
        {
            // OK
        }
        else
        {
            // NOK
        }
    });
}
catch (Exception error)
{
    // Log
}

edited Mar 30 '16 at 14:12

Sam

6,961
15
44
63

answered Jul 09 '12 at 18:47

David

1,559
10
21

2

string jsonToSend = JsonHelper.ToJson(json); Can you explain this line? – Emre Kenci Jun 07 '13 at 12:40
it just convert the object to an json string. (json = object, jsonToSend = json representation of "json"). I should change those names. – David Jun 25 '13 at 13:37
How to attach file to your request? – OPV Jul 23 '17 at 19:46

score 16 · Answer 2 · answered Oct 15 '13 at 07:59

16

This way works fine for me:

var request = new RestSharp.RestRequest("RESOURCE", RestSharp.Method.POST) { RequestFormat = RestSharp.DataFormat.Json }
                .AddBody(BODY);

var response = Client.Execute(request);

// Handle response errors
HandleResponseErrors(response);

if (Errors.Length == 0)
{ }
else
{ }

Hope this helps! (Although it is a bit late)

answered Oct 15 '13 at 07:59

Kutyel

6,297
2
26
55

Yours is the only one that worked for my node.js API. Thanks! – coffekid Jan 15 '14 at 15:02
5

you did not define Client – Edwin O. Feb 09 '16 at 15:26

score 8 · Answer 3 · answered Jul 24 '17 at 07:30

As of 2017 I post to a rest service and getting the results from it like that:

        var loginModel = new LoginModel();
        loginModel.DatabaseName = "TestDB";
        loginModel.UserGroupCode = "G1";
        loginModel.UserName = "test1";
        loginModel.Password = "123";

        var client = new RestClient(BaseUrl);

        var request = new RestRequest("/Connect?", Method.POST);
        request.RequestFormat = DataFormat.Json;
        request.AddBody(loginModel);

        var response = client.Execute(request);

        var obj = JObject.Parse(response.Content);

        LoginResult result = new LoginResult
        {
            Status = obj["Status"].ToString(),
            Authority = response.ResponseUri.Authority,
            SessionID = obj["SessionID"].ToString()
        };

As of now/5/28/2019 you would use AddJsonBody instead of AddBody since you have chosen DataFormat.Json instead of XML. This does both object to JSON and adding to the body of the request in one statement. — Chris, May 28 '19 at 14:30

score 2 · Answer 4 · answered Apr 09 '20 at 10:18

it is better to use json after post your resuest like below

  var clien = new RestClient("https://smple.com/");
  var request = new RestRequest("index", Method.POST);
  request.AddHeader("Sign", signinstance);    
  request.AddJsonBody(JsonConvert.SerializeObject(yourclass));
  var response = client.Execute<YourReturnclassSample>(request);
  if (response.StatusCode == System.Net.HttpStatusCode.Created)
   {
       return Ok(response.Content);
   }

crabCRUSHERclamCOLLECTOR · Answer 5 · 2018-04-10T14:53:35.643

I added this helper method to handle my POST requests that return an object I care about.

For REST purists, I know, POSTs should not return anything besides a status. However, I had a large collection of ids that was too big for a query string parameter.

Helper Method:

    public TResponse Post<TResponse>(string relativeUri, object postBody) where TResponse : new()
    {
        //Note: Ideally the RestClient isn't created for each request. 
        var restClient = new RestClient("http://localhost:999");

        var restRequest = new RestRequest(relativeUri, Method.POST)
        {
            RequestFormat = DataFormat.Json
        };

        restRequest.AddBody(postBody);

        var result = restClient.Post<TResponse>(restRequest);

        if (!result.IsSuccessful)
        {
            throw new HttpException($"Item not found: {result.ErrorMessage}");
        }

        return result.Data;
    }

Usage:

    public List<WhateverReturnType> GetFromApi()
    {
        var idsForLookup = new List<int> {1, 2, 3, 4, 5};

        var relativeUri = "/api/idLookup";

        var restResponse = Post<List<WhateverReturnType>>(relativeUri, idsForLookup);

        return restResponse;
    }

RestRequest.AddBody(object) is obsolete: Use AddXmlBody or AddJsonBody — Luca Ziegler, Feb 04 '20 at 11:15

How to POST request using RestSharp

5 Answers5

Linked