Question

我尝试使用RestSharp客户端发送请求，如下所示我将验证码传递给以下函数

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();
    }
}

但我得到了回复.StatusCode = Bad Request。任何人都可以帮助我如何使用Restsharp客户端发布请求。

Answer 1

我的RestSharp POST 方法：

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
}

Answer 2

这种方式适合我：

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
{ }

希望这有帮助！（虽然有点晚了）

Answer 3

截至2017年，我发布了休息服务并从中得到结果：

        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()
        };

Answer 4

像下面这样发布您的需求后最好使用json

  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);
   }

Answer 5

我添加了这个帮助器方法来处理返回我关心的对象的POST请求。

对于REST纯粹主义者，我知道，除了状态之外，POST不应返回任何内容。但是，我有一大堆id，它们对于查询字符串参数来说太大了。

助手方法：

    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;
    }

用法：

    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;
    }

如何使用RestSharp发送请求

5 个答案: