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.

Sam · Accepted Answer · 2016-03-30 14:12:30Z

81

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, 2016 at 14:12

Sam

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answered Jul 9, 2012 at 18:47

David

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5 Comments

Emre K. Over a year ago

string jsonToSend = JsonHelper.ToJson(json); Can you explain this line?

David Over a year ago

it just convert the object to an json string. (json = object, jsonToSend = json representation of "json"). I should change those names.

POV Over a year ago

How to attach file to your request?

Yster Over a year ago

Seems JsonHelper.ToJson() no longer available.

robino16 Over a year ago

In my case I had to use "application/json" without the "charset=utf-8" part.

Kutyel · Accepted Answer · 2013-10-15 07:59:20Z

22

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, 2013 at 7:59

Kutyel

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2 Comments

coffekid Over a year ago

Yours is the only one that worked for my node.js API. Thanks!

Edwin O. Over a year ago

you did not define Client

Mehmet Recep Yildiz · Accepted Answer · 2017-07-24 07:30:52Z

11

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()
        };

answered Jul 24, 2017 at 7:30

Mehmet Recep Yildiz

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1 Comment

Chris Over a year ago

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.

Mahbobeh Porniazi · Accepted Answer · 2020-04-09 10:18:46Z

8

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);
   }

answered Apr 9, 2020 at 10:18

Mahbobeh Porniazi

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Comments

crabCRUSHERclamCOLLECTOR · Accepted Answer · 2018-04-10 14:53:35Z

2

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;
    }

edited Apr 10, 2018 at 14:53

answered Apr 10, 2018 at 14:40

crabCRUSHERclamCOLLECTOR

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1 Comment

Luca Ziegler Over a year ago

RestRequest.AddBody(object) is obsolete: Use AddXmlBody or AddJsonBody

Deepak · Accepted Answer · 2022-05-30 09:15:23Z

0

It's better to specify the Json data body.

var client = new RestClient("URL");
var request = new RestRequest("Resources",Method.POST);
request.RequestFormat = RestSharp.DataFormat.Json;
request.AddBody(new classname
{
  id = value1;
  Name = "";

});
var response = client.Execute(request);

Check statusCode from response.

Always ensure that status code should be OK.

edited May 30, 2022 at 9:15

answered May 30, 2022 at 9:14

Deepak

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