Question

您好我正在使用以下方法发布到网址。这对我来说可以。我在这里创造了五个任务。

private async Task CreateMultipleTasksAsync(string url, ExtendedWebClient oExtendedWebClient, string sParam)
        {

            try
            {
                Task<string> download1 = oExtendedWebClient.ProcessURLAsync(url, oExtendedWebClient, sParam);
                Task<string> download2 = oExtendedWebClient.ProcessURLAsync(url, oExtendedWebClient, sParam);
                Task<string> download3 = oExtendedWebClient.ProcessURLAsync(url, oExtendedWebClient, sParam);
                Task<string> download4 = oExtendedWebClient.ProcessURLAsync(url, oExtendedWebClient, sParam);
                Task<string> download5 = oExtendedWebClient.ProcessURLAsync(url, oExtendedWebClient, sParam);
                lst_tasks1.Add(download1);
                lst_tasks2.Add(download2);
                lst_tasks3.Add(download3);
                lst_tasks4.Add(download4);
                lst_tasks5.Add(download5);

                // Await each task.
                 Result1 = await download1;
                 Result2 = await download2;
                 Result3 = await download3;
                 Result4 = await download4;
                 Result5 = await download5;

            }
            catch (Exception ex)
            {
                ErrorLog.createLog("ex.StackTrace = " + ex.StackTrace + " ex.tostring = " + ex.ToString());
            }

        }

我需要实现的是，如果任何任务返回包含＆＃34; START＆＃34;字停止等待并继续..

我这样做的方式如下：

   Task.WaitAll(lst_tasks1.ToArray());
if (Result1.ToLower().Contains("START") && !Result1.Contains(sTextSearch))
{
    goto call2;
}
else
{

worker.ReportProgress(0, "Waiting for tasks2");
Task.WaitAll(lst_tasks2.ToArray());

依旧等待任务5. 有没有办法使用Task.waitany来做这个代码请建议

Answer 1

You could try using the solution given in this post

您需要将条件作为谓词传递给此方法。

假设lst_tasks包含所有任务，你可以这样做。

await WhenAny(lst_tasks, s => s == "START");

一旦任务返回START，这将停止等待。

需要帮助等待异步webclient帖子响应

1 个答案: