4

responseObject我有一个这样的方法，在使用 Task.Factory.FromAsync()声明时挂起

private async Task<string> post(string url, string postdata)
        {
            var request = WebRequest.Create(new Uri(url)) as HttpWebRequest;
            request.Method = "POST";

            // this works
            Task<Stream> requestStream = Task<Stream>.Factory.FromAsync(request.BeginGetRequestStream, request.EndGetRequestStream, request);
            var sw = new StreamWriter(requestStream.Result);
            byte[] data = Encoding.UTF8.GetBytes(postdata);
            await requestStream.Result.WriteAsync(data, 0, data.Length);

            // this hangs
            Task<WebResponse> responseObject = Task<WebResponse>.Factory.FromAsync(request.BeginGetResponse, request.EndGetResponse, request); // Hangs here
            // Doesn't get to here
            var responseStream = responseObject.Result.GetResponseStream();
            var sr = new StreamReader(responseStream);
            string received = await sr.ReadToEndAsync();

            return received;
}

如果我手动运行 Begin/End 方法，它工作正常，如下所示：

        request.BeginGetRequestStream(async ar =>
        {
            var requestStream = request.EndGetRequestStream(ar);
            using (var sw = new StreamWriter(requestStream))
            {
                byte[] data = Encoding.UTF8.GetBytes(postdata);
                await requestStream.WriteAsync(data, 0, data.Length);
            }

            request.BeginGetResponse(async a =>
            {
                HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(a);
                var responseStream = response.GetResponseStream();
                using (var sr = new StreamReader(responseStream))
                {
                    string received = await sr.ReadToEndAsync();
                }

            }, null);
        }, null);

但是，在这种情况下，ReadToEndAsync() 在单独的线程上运行，并且我希望该方法返回结果，这是不可能的，因为该方法在 BeginGetResponse 回调完成之前返回。

在这一点上，我认为我做的非常错误并且完全错误的方式，所以任何帮助将不胜感激。

4

1 回答 1

11

首先：不要在异步方法中使用“.Result”。这将阻止正在运行您的方法的线程。请改用“await”，以便在获取结果时线程跳回您的方法。

您的代码的问题是您打开了一个请求流，但您从不关闭它。那么它应该什么时候完成您的请求并发送它呢？它总是期望有更多的输入，直到请求超时运行。此外，您还没有设置内容类型和内容长度。

private async Task<string> post(string url, string postdata)
    {
        var request = WebRequest.Create(new Uri(url)) as HttpWebRequest;
        request.Method = "POST";

        byte[] data = Encoding.UTF8.GetBytes(postdata);
        request.ContentLength = data.Length;

        using (var requestStream = await Task<Stream>.Factory.FromAsync(request.BeginGetRequestStream, request.EndGetRequestStream, request))
        {
            await requestStream.WriteAsync(data, 0, data.Length);
        }


        WebResponse responseObject = await Task<WebResponse>.Factory.FromAsync(request.BeginGetResponse, request.EndGetResponse, request);
        var responseStream = responseObject.GetResponseStream();
        var sr = new StreamReader(responseStream);
        string received = await sr.ReadToEndAsync();

        return received;
}
于 2012-12-31T07:45:38.623 回答