如何使普通的WebRequest异步和可唤醒

本文关键字：唤醒 异步 WebRequest 何使普 | 更新日期: 2023-09-27 18:27:33

我需要使以下代码异步且可唤醒。

我需要从web服务器获取大量数据，然后这些数据将用于填充我的应用程序中的xaml页面。

因此，我需要DefLogin（）方法是不可用的。

有可能吗？

public void DefLogin()
    {
        postData = "My Data To Post";
        var url = new Uri("Url To Post to", UriKind.Absolute);
        webRequest = WebRequest.Create(url);
        webRequest.Method = "POST";
        webRequest.ContentType = "text/xml";
        webRequest.BeginGetRequestStream(new AsyncCallback(GetRequestStreamCallback), webRequest);
    }
    public void GetRequestStreamCallback(IAsyncResult asynchronousResult)
    {
        webRequest = (HttpWebRequest)asynchronousResult.AsyncState;
        Stream postStream = webRequest.EndGetRequestStream(asynchronousResult);
        byte[] byteArray = Encoding.UTF8.GetBytes(postData);
        postStream.Write(byteArray, 0, byteArray.Length);
        postStream.Close();
        Debug.WriteLine("Start BEGINGetResponse");
        webRequest.BeginGetResponse(new AsyncCallback(GetResponseCallback), webRequest);
    }
    public void GetResponseCallback(IAsyncResult asynchronousResult)
    {
        try
        {
            HttpWebRequest webRequest = (HttpWebRequest)asynchronousResult.AsyncState;
            HttpWebResponse response;
            response = (HttpWebResponse)webRequest.EndGetResponse(asynchronousResult);
            Stream streamResponse = response.GetResponseStream();
            StreamReader streamReader = new StreamReader(streamResponse);
            string Response = streamReader.ReadToEnd();
            streamResponse.Close();
            streamReader.Close();
            response.Close();
            if (Response == "")
            {
                //show some error msg to the user        
                Debug.WriteLine("ERROR");
            }
            else
            {
                //Your response will be available in "Response" 
                Debug.WriteLine(Response);
            }
        }
        catch (WebException)
        {
            //error    
        }
    }

我在StackOverflow上看到了这个问题：用Async和Await转换普通的Http-Post web请求，但我无法正确理解答案。

有人能帮忙吗？我真的很感激！

如何使普通的WebRequest异步和可唤醒

您可以使用TaskFactory.FromAsync将APM转换为TAP，生成许多像这样的微小扩展方法：

public static Task<Stream> GetRequestStreamAsync(this WebRequest request)
{
  return TaskFactory.FromAsync(request.BeginGetRequestStream, request.EndGetRequestStream, null);
}

并对WebRequest.GetResponse和（如有必要）Stream.Write、Stream.Flush等进行相同操作。

然后，您可以使用async和await编写实际逻辑，而无需任何回调：

public async Task DefLoginAsync()
{
    postData = "My Data To Post";
    var url = new Uri("Url To Post to", UriKind.Absolute);
    webRequest = WebRequest.Create(url);
    webRequest.Method = "POST";
    webRequest.ContentType = "text/xml";
    using (Stream postStream = await webRequest.GetRequestStreamAsync())
    {
        byte[] byteArray = Encoding.UTF8.GetBytes(postData);
        await postStream.WriteAsync(byteArray, 0, byteArray.Length);
        await postStream.FlushAsync();
    }
    try
    {
        string Response;
        using (var response = (HttpWebResponse)await webRequest.GetResponseAsync());
        using (Stream streamResponse = response.GetResponseStream())
        using (StreamReader streamReader = new StreamReader(streamResponse))
        {
            Response = await streamReader.ReadToEndAsync();
        }
        if (Response == "")
        {
            //show some error msg to the user        
            Debug.WriteLine("ERROR");
        }
        else
        {
            //Your response will be available in "Response" 
            Debug.WriteLine(Response);
        }
    }
    catch (WebException)
    {
        //error    
    }
}
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