MSDN有一个样本。

using System;
using System.IO;
using System.Net;
using System.Text;

namespace Examples.System.Net
{
    public class WebRequestPostExample
    {
        public static void Main()
        {
            // Create a request using a URL that can receive a post. 
            WebRequest request = WebRequest.Create("http://www.contoso.com/PostAccepter.aspx");
            // Set the Method property of the request to POST.
            request.Method = "POST";
            // Create POST data and convert it to a byte array.
            string postData = "This is a test that posts this string to a Web server.";
            byte[] byteArray = Encoding.UTF8.GetBytes(postData);
            // Set the ContentType property of the WebRequest.
            request.ContentType = "application/x-www-form-urlencoded";
            // Set the ContentLength property of the WebRequest.
            request.ContentLength = byteArray.Length;
            // Get the request stream.
            Stream dataStream = request.GetRequestStream();
            // Write the data to the request stream.
            dataStream.Write(byteArray, 0, byteArray.Length);
            // Close the Stream object.
            dataStream.Close();
            // Get the response.
            WebResponse response = request.GetResponse();
            // Display the status.
            Console.WriteLine(((HttpWebResponse)response).StatusDescription);
            // Get the stream containing content returned by the server.
            dataStream = response.GetResponseStream();
            // Open the stream using a StreamReader for easy access.
            StreamReader reader = new StreamReader(dataStream);
            // Read the content.
            string responseFromServer = reader.ReadToEnd();
            // Display the content.
            Console.WriteLine(responseFromServer);
            // Clean up the streams.
            reader.Close();
            dataStream.Close();
            response.Close();
        }
    }
}

如果你喜欢一个流畅的API，你可以使用Tiny.RestClient。在NuGet上可以买到。

var client = new TinyRestClient(new HttpClient(), "http://MyAPI.com/api");
// POST
var city = new City() { Name = "Paris", Country = "France" };
// With content
var response = await client.PostRequest("City", city)
                           .ExecuteAsync<bool>();

到目前为止，我找到了简单的解决方案(一行程序，没有错误检查，没有等待响应):

(new WebClient()).UploadStringAsync(new Uri(Address), dataString);‏

请谨慎使用!

这是一个HTTPS web请求的例子。可以在PHP脚本中回显任何结果。最后，PHP回显字符串将在c#客户端显示为警报。

string url = "https://mydomain.ir/test1.php";
StringBuilder postData = new StringBuilder();
postData.Append(String.Format("{0}={1}&", HttpUtility.HtmlEncode("username"), HttpUtility.HtmlEncode("ali")));
postData.Append(String.Format("{0}={1}", HttpUtility.HtmlEncode("password"), HttpUtility.HtmlEncode("123456789")));
StringContent myStringContent = new StringContent(postData.ToString(), Encoding.UTF8, "application/x-www-form-urlencoded");
HttpClient client = new HttpClient();
HttpResponseMessage message = client.PostAsync(url, myStringContent).GetAwaiter().GetResult();
string responseContent = message.Content.ReadAsStringAsync().GetAwaiter().GetResult();

DisplayAlert("Your Feedback", responseContent, "OK");

PHP服务器端:

<?php
  if (isset($_POST["username"]) && $_POST["username"] == "ali") {
    echo "Yes, hi Ali";
  }
  else {
    echo "No, where is Ali?";
  }
?>

结果将是“Yes, hi Ali”。

这是为Xamarin形式。对于一个c# .NET应用程序，将DisplayAlert替换为:

MessageBox.show(responseContent);

当使用Windows.Web.Http命名空间时，对于POST而不是FormUrlEncodedContent，我们编写HttpFormUrlEncodedContent。同样，响应类型为HttpResponseMessage。其余的就像Evan Mulawski写的那样。

c# . net

    using System.Net.Http;
    
    private static readonly HttpClient httpClient = new HttpClient();

//POST    
    var values = new Object();
    values[0] = "Value1";
    values[2] = "Value2";
    values[n] = "ValueN";

    var content = new FormUrlEncodedContent(values);
    var response = await httpClient.PostAsync("URL", content);
    var responseString = await response.Content.ReadAsStringAsync();

    

//GET
 var response = await httpClient.GetStringAsync("URL");