使用 解析xml为json 并对json进行处理

摘要：  string jieguo="json字符串";   XmlDocument doc = new XmlDocument();     &nb…

  string jieguo="json字符串";
  XmlDocument doc = new XmlDocument();
                    doc.LoadXml(jieguo);
                    string json = Newtonsoft.Json.JsonConvert.SerializeXmlNode(doc);
                    var chulixinxi = JsonConvert.DeserializeObject(json) as JObject;
                    string chulixinxitype = chulixinxi.Last.Path;
                    if (chulixinxitype == "error_response")
                    {
                        jieguoshuchu = mobilephone + "|" + chulixinxi["error_response"]["sub_msg"].ToString();
                    }
                    if (chulixinxitype == "dingtalk_corp_ext_add_response")
                    {
                        jieguoshuchu = mobilephone + "|添加成功";
                    }