异步请求Url

 public static System.Xml.Linq.XElement Post(HttpClient _myhttp,   string url, XmlDocument doc)
        {          
            HttpContent content = new StringContent(doc.InnerXml, Encoding.UTF8, "application/xml");
            //Task 开启一个线程，然后Run起来
             //(可以传参数)=>{逻辑代码}
            var message = Task<HttpResponseMessage>.Run<HttpResponseMessage>(() =>
            {
                return _myhttp.PostAsync(url, content);
            });
            message.Wait();
            //接收返回得信息
            if (message.Result.IsSuccessStatusCode)
            {
                var s = Task<string>.Run<string>(() =>
                {
                    return message.Result.Content.ReadAsStringAsync();
                });
                s.Wait();
                //最后是把S.Result转成XElement的形式
                return System.Xml.Linq.XElement.Load(s.Result);
            }
            else
            {
                throw new Exception("StatusCode:" + message.Result.StatusCode.ToString());
            }
        }