C#做一个Http协议接口

下面是一个简单的C# Http协议接口示例，可以响应GET和POST请求：

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

class HttpServer {
    public static void Main(string[] args) {
        HttpListener listener = new HttpListener();
        listener.Prefixes.Add("http://localhost:8080/"); // 监听的地址和端口号
        listener.Start();
        Console.WriteLine("Listening...");

        while (true) {
            // 等待请求
            HttpListenerContext context = listener.GetContext();
            HttpListenerRequest request = context.Request;
            HttpListenerResponse response = context.Response;

            Console.WriteLine("{0} {1}", request.HttpMethod, request.Url);

            // 处理GET请求
            if (request.HttpMethod == "GET") {
                string responseString = "Hello World!";
                byte[] buffer = Encoding.UTF8.GetBytes(responseString);
                response.ContentLength64 = buffer.Length;
                Stream output = response.OutputStream;
                output.Write(buffer, 0, buffer.Length);
                output.Close();
            }

            // 处理POST请求
            else if (request.HttpMethod == "POST") {
                StreamReader reader = new StreamReader(request.InputStream);
                string requestBody = reader.ReadToEnd();
                Console.WriteLine("Request Body: {0}", requestBody);
                string responseString = "Received Request: " + requestBody;
                byte[] buffer = Encoding.UTF8.GetBytes(responseString);
                response.ContentLength64 = buffer.Length;
                Stream output = response.OutputStream;
                output.Write(buffer, 0, buffer.Length);
                output.Close();
            }
        }
    }
}

在上面的示例中，我们监听了地址和端口号为`http://localhost:8080/`的请求。当接收到GET请求时，返回"Hello World!"；当接收到POST请求时，返回收到的请求体。你可以根据实际需求修改响应的内容和处理方法。