使用简单客户端。UploadString(地址、内容);正常工作很好，但我认为应该记住，如果没有返回HTTP成功状态码，将抛出webeexception。我通常这样处理，打印远程服务器返回的任何异常消息:

try
{
    postResult = client.UploadString(address, content);
}
catch (WebException ex)
{
    String responseFromServer = ex.Message.ToString() + " ";
    if (ex.Response != null)
    {
        using (WebResponse response = ex.Response)
        {
            Stream dataRs = response.GetResponseStream();
            using (StreamReader reader = new StreamReader(dataRs))
            {
                responseFromServer += reader.ReadToEnd();
                _log.Error("Server Response: " + responseFromServer);
            }
        }
    }
    throw;
}

使用WebClient。UploadString或WebClient。UploadData可以轻松地将数据POST到服务器。我将展示一个使用UploadData的示例，因为UploadString的使用方式与DownloadString相同。

byte[] bret = client.UploadData("http://www.website.com/post.php", "POST",
                System.Text.Encoding.ASCII.GetBytes("field1=value1&field2=value2") );
 
string sret = System.Text.Encoding.ASCII.GetString(bret);

更多:http://www.daveamenta.com/2008-05/c-webclient-usage/

下面是简单明了的答案:

public String sendSMS(String phone, String token) {
    WebClient webClient = WebClient.create(smsServiceUrl);

    SMSRequest smsRequest = new SMSRequest();
    smsRequest.setMessage(token);
    smsRequest.setPhoneNo(phone);
    smsRequest.setTokenId(smsServiceTokenId);

    Mono<String> response = webClient.post()
          .uri(smsServiceEndpoint)
          .header(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
          .body(Mono.just(smsRequest), SMSRequest.class)
          .retrieve().bodyToMono(String.class);

    String deliveryResponse = response.block();
    if (deliveryResponse.equalsIgnoreCase("success")) {
      return deliveryResponse;
    }
    return null;
}

我刚刚找到了解决方案，是的，它比我想象的要简单:)

这就是解决方案:

string URI = "http://www.myurl.com/post.php";
string myParameters = "param1=value1&param2=value2&param3=value3";

using (WebClient wc = new WebClient())
{
    wc.Headers[HttpRequestHeader.ContentType] = "application/x-www-form-urlencoded";
    string HtmlResult = wc.UploadString(URI, myParameters);
}

它很有魅力:)

有一个内置的方法叫做UploadValues，它可以发送HTTP POST(或任何类型的HTTP方法)，并以适当的数据格式处理请求体的构造(用“&”连接参数并通过url编码转义字符):

using(WebClient client = new WebClient())
{
    var reqparm = new System.Collections.Specialized.NameValueCollection();
    reqparm.Add("param1", "<any> kinds & of = ? strings");
    reqparm.Add("param2", "escaping is already handled");
    byte[] responsebytes = client.UploadValues("http://localhost", "POST", reqparm);
    string responsebody = Encoding.UTF8.GetString(responsebytes);
}

使用webapiclient与模型发送序列化json参数请求。

PostModel.cs

    public string Id { get; set; }
    public string Name { get; set; }
    public string Surname { get; set; }
    public int Age { get; set; }

WebApiClient.cs

internal class WebApiClient  : IDisposable
  {

    private bool _isDispose;

    public void Dispose()
    {
        Dispose(true);
        GC.SuppressFinalize(this);
    }

    public void Dispose(bool disposing)
    {
        if (!_isDispose)
        {

            if (disposing)
            {

            }
        }

        _isDispose = true;
    }

    private void SetHeaderParameters(WebClient client)
    {
        client.Headers.Clear();
        client.Headers.Add("Content-Type", "application/json");
        client.Encoding = Encoding.UTF8;
    }

    public async Task<T> PostJsonWithModelAsync<T>(string address, string data,)
    {
        using (var client = new WebClient())
        {
            SetHeaderParameters(client);
            string result = await client.UploadStringTaskAsync(address, data); //  method:
    //The HTTP method used to send the file to the resource. If null, the default is  POST 
            return JsonConvert.DeserializeObject<T>(result);
        }
    }
}

业务调用方法

    public async Task<ResultDTO> GetResultAsync(PostModel model)
    {
        try
        {
            using (var client = new WebApiClient())
            {
                var serializeModel= JsonConvert.SerializeObject(model);// using Newtonsoft.Json;
                var response = await client.PostJsonWithModelAsync<ResultDTO>("http://www.website.com/api/create", serializeModel);
                return response;
            }
        }
        catch (Exception ex)
        {
            throw new Exception(ex.Message);
        }

    }