摘要
在网络传输过程中，不能保证所有的请求都能正确的被服务端接受或者处理，那么进行简单的重试可以进行简单的补救。比如现在大部分支付功能，在支付成功之后，需要回调我们网站的接口，并且要求我们的接口给一个类似成功的响应，支付端收到我们的成功响应，就不再进行重试请求了。
比如：支付宝文档中有这么一段描述
Polly简单实用
???????private ILog _log; ???????/// <summary> ???????/// 重试时间点 ???????/// </summary> ???????private TimeSpan[] _retryTimes = new[] { ???????????TimeSpan.FromSeconds(1), ???????????TimeSpan.FromSeconds(5), ???????????TimeSpan.FromSeconds(10), ???????????TimeSpan.FromSeconds(20), ???????????TimeSpan.FromMinutes(1) ???????}; ???????/// <summary> ???????/// 提交文本 ???????/// </summary> ???????/// <param name="url"></param> ???????/// <param name="data"></param> ???????/// <returns></returns> ???????public string PostTxt(string url, Dictionary<string, string> paras) ???????{ ???????????try ???????????{ ?
????????????????????string data = ConvertDic2QueryString(paras); ????????????????????byte[] byteArray = Encoding.UTF8.GetBytes(data);
???????????????var retry4TimePolicy = Polly.Policy.Handle<WebException>().WaitAndRetry( ?????????????????_retryTimes, ?????????????????????????(ex, ts, i, context) => ?????????????????????????{ ?????????????????????????????_log.Info(new LogModel ?????????????????????????????{ ?????????????????????????????????Content = $"Request:{url},data:{ConvertDic2QueryString(paras)},第{i}次重试", ?????????????????????????????????userId= "", ?????????????????????????????????Op = "posttxt" ?????????????????????????????}); ?????????????????????????}); ???????????????return retry4TimePolicy.Execute(() => ????????????????{ ???????????????????????????????????????HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url); ????????????????????request.ContentType = "application/x-www-form-urlencoded"; ????????????????????request.Method = "POST"; ????????????????????request.ContentLength = byteArray.Length; ????????????????????request.Timeout = 60000; ????????????????????if (url.StartsWith("https")) ????????????????????{ ????????????????????????request.ServerCertificateValidationCallback += (sender, certificate, chain, sslPolicyErrors) => true; ????????????????????} ????????????????????Stream requestStream = request.GetRequestStream(); ????????????????????requestStream.Write(byteArray, 0, byteArray.Length); ????????????????????HttpWebResponse webResponse = (HttpWebResponse)request.GetResponse(); ????????????????????StreamReader reader = new StreamReader(webResponse.GetResponseStream(), Encoding.UTF8); ????????????????????String sResult = reader.ReadToEnd(); ????????????????????requestStream.Close(); ????????????????????reader.Close(); ????????????????????webResponse.Close(); ????????????????????return sResult; ????????????????}); ???????????} ???????????catch (WebException ex) ???????????{ ???????????????throw ex; ???????????} ??????????????????} ?public string ConvertDic2QueryString(Dictionary<string, string> dic) ???????{ ???????????if (dic == null) ???????????{ ???????????????return string.Empty; ???????????} ???????????StringBuilder sb = new StringBuilder(); ???????????foreach (var key in dic.Keys) ???????????{ ???????????????sb.AppendFormat("{0}={1}&", key, HttpUtility.UrlEncode(dic[key])); ???????????} ???????????return sb.ToString().TrimEnd(‘&‘); ???????}可以类似上面的一种请求，在捕获到webException异常时，立马重试一次，检查网络是否正常，如果仍不能正确进行请求，下次5s，10s，20s，进行尝试，如果仍不正常，则在1分钟后进行最后尝试。否则认为网络不正常，则抛出异常，进行日志记录。
总结
Polly支持多种重试策略，感兴趣的可以移步这里。
https://www.cnblogs.com/CreateMyself/p/7589397.html
http://www.cnblogs.com/xishuai/p/asp-net-core-polly.html
[Asp.net core]使用Polly网络请求异常重试
原文地址：https://www.cnblogs.com/wolf-sun/p/8761380.html