Question

我希望异步运行Windows服务中长时间运行的进程 - 每隔3秒轮询一次进程并使用SignalR报告。下面的代码（理论上）将以基于事件的方式执行此操作，但我不想在每次更改时触发进度。

有人可以提供一个简单的示例，说明如何专门实现此过程以启动流程和轮询/报告进度。请记住，我已经完成了几年的全职开发！

public async Task<string> StartTask(int delay)
{
    var tokenSource = new CancellationTokenSource();
    var progress = new Progress<Tuple<Int32, Int32>>();

    progress.ProgressChanged += (s, e ) =>
    {
        r2hProcessesProxy.Invoke("DownloadNotify", string.Format("Major={0} - Minor={1}", e.Item1 , e.Item2  ));
    };

    var task = DownloadTask.DoDownload(delay, tokenSource.Token, new Progress<Tuple<Int32,Int32>>(p => new Tuple<Int32, Int32>(0,0)));

    await task;

    return "Task result";
}

Answer 1

您可以使用Reactive Extensions（Rx）执行此操作。查看Throttle()方法：https://msdn.microsoft.com/en-us/library/hh229400(v=vs.103).aspx

using System;
using System.Linq;
using System.Reactive.Linq;
using System.Threading.Tasks;

public class Test
{
    public async Task<string> StartTask(int delay)
    {
        var tokenSource = new CancellationTokenSource();
        var progress = new Progress<Tuple<Int32, Int32>>();

        var observable = Observable.FromEvent<EventHandler<Tuple<Int32, Int32>>, Tuple<Int32, Int32>>(h => progress.ProgressChanged += h, h => progress.ProgressChanged -= h);
        var throttled = observable.Throttle(TimeSpan.FromSeconds(3));

        using (throttled.Subscribe(e =>
        {
            r2hProcessesProxy.Invoke("DownloadNotify", string.Format("Major={0} - Minor={1}", e.Item1, e.Item2));
        }))
        {
            await DownloadTask.DoDownload(delay, tokenSource.Token, new Progress<Tuple<Int32, Int32>>(p => new Tuple<Int32, Int32>(0, 0)));
        }

        return "result";
    }
}

检查http://go.microsoft.com/fwlink/?LinkId=208528以获取有关RX

的更多信息

.NET 4.5轮询进度示例请

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