使用SimpleHTTPServer将数据发送回HTML

时间:2016-06-21 10:52:01

标签: python html post get server

我有一个表单,它在一个模仿登录操作的按钮上有一个GET操作

 public DataTable readXML(string postcode, string response, string accessCode)
    {
        //Create URL
        string url = $"http://pcls1.craftyclicks.co.uk/xml/rapidaddress?postcode={postcode}&response={response}&key={accessCode}";

        try
        {
            DataTable dtReturn = new DataTable();
            dtReturn.Columns.Add("PropertyItem", Type.GetType("System.String"));
            dtReturn.Columns.Add("PropertyValue", Type.GetType("System.String"));

            //Create WebRequest
            WebRequest request = WebRequest.Create(url);
            using (Stream responseStream = request.GetResponse().GetResponseStream())
            {
                if (responseStream != null)
                {
                    using (TextReader textReader = new StreamReader(responseStream))
                    {
                        XmlTextReader reader = new XmlTextReader(textReader);
                        postcode = postcode.Replace(" ", "");

                        string option1 = "";
                        string option2 = "";

                        while (reader.Read())
                        {

                            if (reader.Name.Equals("line_1"))
                            {
                                option1 = reader.ReadString();
                            }
                            if (reader.Name.Equals("udprn"))
                            {
                                option2 = reader.ReadString();
                            }
                            if (option1 != "" && option2 != "")
                            {
                                dtReturn.Rows.Add(new object[] { option1, option2 });
                                option1 = "";
                                option2 = "";
                            }
                        }
                        return dtReturn;
                    }
                }
                throw new Exception("ResponseStream is NULL");
            }
        }
        catch (WebException ex)
        {
            return null;
        }
    }

我已在服务器端实现了do_GET方法来处理此请求并获取令牌

Socket client = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);

我不想导航到提交的新页面 如何将令牌值传递回我的html代码以显示登录成功?例如作为新标签

我对HTML很新,任何帮助都会受到赞赏!

0 个答案:

没有答案
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