c# – 如何从方法外部获得异步Web请求的响应?
发布时间:2020-12-15 17:24:53  所属栏目:百科  来源:网络整理 
            导读:我有点困惑.我试图以异步方式发布到我的Web服务,理想情况下我想启动请求,在UI上显示加载微调器,然后当异步请求完成处理响应时,如果有响应则显示错误,或者用结果做另一个操作. 这是我的代码,我在这里调用请求并传递一些数据. private void SignInExecute(){ i
                
                
                
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 我有点困惑.我试图以异步方式发布到我的Web服务,理想情况下我想启动请求,在UI上显示加载微调器,然后当异步请求完成处理响应时,如果有响应则显示错误,或者用结果做另一个操作. 
  
  这是我的代码,我在这里调用请求并传递一些数据. private void SignInExecute()
{
        if (Username == null || Password == null)
        {
            LoginOutput = "Please provide a username or password.";
        }
        else
        {
            this.webService.SendLoginRequest("http://localhost:3000/client_sessions","username=" + Username + "&password=" + Password);
        }
}这是实际的Web请求代码: public void SendLoginRequest(string url,string postdata)
{
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
        request.Method = "POST";
        request.ContentType = "application/x-www-form-urlencoded";
        request.Accept = "application/json";
        byte[] byteArray = Encoding.UTF8.GetBytes(postdata);
        request.CookieContainer = new CookieContainer();
        request.ContentLength = byteArray.Length;
        Stream dataStream = request.GetRequestStream();
        dataStream.Write(byteArray,byteArray.Length);
        dataStream.Close();
        ((HttpWebRequest)request).KeepAlive = false;
        request.BeginGetResponse(new AsyncCallback(GetLoginResponseCallback),request);
    }
    private static void GetLoginResponseCallback(IAsyncResult asynchronousResult)
    {
        HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;
        // End the operation
        HttpWebResponse response =  (HttpWebResponse)request.EndGetResponse(asynchronousResult);
        Stream streamResponse = response.GetResponseStream();
        StreamReader streamRead = new StreamReader(streamResponse);
        string responseString = streamRead.ReadToEnd();
        Console.WriteLine(responseString);
        // Close the stream object
        streamResponse.Close();
        streamRead.Close();
        response.Close();
    }所以总结一下.我希望能够将响应返回给最初调用Web请求启动的对象.有帮助吗? 解决方法
 您需要告诉BeginGetResponse返回到通过SynchronizationContext.Current调用它的相同上下文.像这样的东西(代码没有正确的错误检查,所以你应该正确考虑)(另外,Platinum Azure是正确的,你应该使用一个使你的流正常关闭(并保证): 
  
  在您的SendLoginRequest中: //Box your object state with the current thread context
object[] boxedItems = new []{request,SynchronizationContext.Current};
request.BeginGetResponse(new AsyncCallback(GetLoginResponseCallback),boxedItems);getresponse代码: private static void GetLoginResponseCallback(IAsyncResult asynchronousResult)
{
//MY UPDATE 
    //Unbox your object state with the current thread context
    object[] boxedItems = asynchronousResult.AsyncState as object[];
    HttpWebRequest request = boxedItems[0] as HttpWebRequest;
    SynchronizationContext context = boxedItems[1] as SynchronizationContext;
    // End the operation
    using(HttpWebResponse response =  
        (HttpWebResponse)request.EndGetResponse(asynchronousResult))
    {
        using(Stream streamResponse = response.GetResponseStream())
        {
            using(StreamReader streamRead = new StreamReader(streamResponse))
            {
                string responseString = streamRead.ReadToEnd();
                Console.WriteLine(responseString);
//MY UPDATE
                //Make an asynchronous call back onto the main UI thread 
                //(context.Send for a synchronous call)
                //Pass responseString as your method parameter 
                //If you have more than one object,you will have to box again
                context.Post(UIMethodToCall,responseString);
            }
        }
    }
}实现UI处理 public static void UIMethodCall(object ObjectState)
{
    String response = ObjectState as String;
    label1.Text = String.Format("Output: {0}",response);
    //Or whatever you need to do in the UI...
}现在,我会先测试一下这个.我对Microsoft实现事件驱动异步的理解是,响应是上下文感知的,并且知道返回哪个上下文.因此,在跳到假设您不在同一个上下文之前,通过尝试更新UI来测试它(如果您不在调用(UI)线程上,这将导致线程上下文异常) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! | 
