Question

void GetResponseCallback(IAsyncResult asynchronousResult)
        {

            try
            {
                HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;
                HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asynchronousResult);
                Stream streamResponse = response.GetResponseStream();
                StreamReader streamRead = new StreamReader(streamResponse);
                string responseString = streamRead.ReadToEnd();

                XmlReader xmlDoc = XmlReader.Create(new MemoryStream(System.Text.UTF8Encoding.UTF8.GetBytes(responseString)));

                while (xmlDoc.Read())
                {
                    if (xmlDoc.NodeType == XmlNodeType.Element)
                    {

                        if (xmlDoc.Name.Equals("ResponseCode"))
                        {
                            responseCode = xmlDoc.ReadInnerXml();

                        }

                    }

                }
                if (Convert.ToInt32(responseCode) == 200)
                {

                   MessageBox.Show("Success");
                }


                // Close the stream object
                streamResponse.Close();
                streamRead.Close();
                // Release the HttpWebResponse
                response.Close();


            }
            catch (WebException e)
            {
                // Error treatment
                // ...
            }
        }

请告诉我如何解决这个问题...

Answer 1

    Dispatcher.BeginInvoke(() =>
            MessageBox.Show("Your message")
    );

Answer 2

从代码中与 UI 进行的任何互动都需要在发件人线索上, 您从 HTTP 请求回调时不会在此线索上运行错误。

你应该可以使用类似

部署 (()\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

以显示消息框

- 鲁珀特。

友情链接