用于建立持久连接的 HTTP 网络请求

Question

如何在同一连接中发出第二个请求？

HttpWebRequest request = (HttpWebRequest)WebRequest.Create("String.url");
WebResponse response = request.GetResponse();
Stream stream = response.GetResponseStream();
StreamReader reader = new StreamReader(stream);
String result = reader.ReadToEnd();
stream.Dispose();
reader.Dispose();

Answer 1

HTTP Web 请求建立持久连接.....您可以同时使用"GET"或"POST"您可以根据需要增加连接(例如 3 到 20 或 50 ...等(

           string webpageContent = "";
           byte[] byteArray = Encoding.UTF8.GetBytes("value");
           HttpWebRequest webRequest (HttpWebRequest)WebRequest.Create(URL);
           webRequest.Method = "POST";
           webRequest.KeepAlive = true;
           webRequest.Timeout = 120000;
           System.Net.ServicePointManager.DefaultConnectionLimit = 3;
           webRequest.ContentType = "application/x-www-form-urlencoded";
           webRequest.ContentLength = byteArray.Length;
           using (Stream webpageStream = webRequest.GetRequestStream())
           {
               webpageStream.Write(byteArray, 0, byteArray.Length);
           }
           using (HttpWebResponse webResponse = (HttpWebResponse)webRequest.GetResponse())
           {
               using (StreamReader reader = new StreamReader(webResponse.GetResponseStream()))
               {
                   webpageContent = reader.ReadToEnd();
               }
           }

Answer 2

HttpWebRequest上的 KeepAlive 属性用于保持连接。它默认为 true .

以下是包含更多详细信息的文档：

https://msdn.microsoft.com/en-us/library/system.net.httpwebrequest.keepalive(v=vs.110(.aspx