Question

由于StandardOutput没有反应，我需要一种方法来观察它。我知道Process类公开了一个事件，用于在写入输出时接收通知所以我使用这个扩展方法来获得标准输出的IObservable

public static class ProcessExtensions
{
    public static IObservable<string> StandardOutputObservable(this Process process)
    {
        process.EnableRaisingEvents = true;
        process.StartInfo.RedirectStandardOutput = true;

        var received = Observable.FromEventPattern<DataReceivedEventHandler,DataReceivedEventArgs>(
            handler => handler.Invoke,
            h => process.OutputDataReceived += h,
            h => process.OutputDataReceived -= h)
            .TakeUntil(Observable.FromEventPattern(
                h => process.Exited += h,
                h => process.Exited -= h))
            .Select(e => e.EventArgs.Data);

        process.BeginOutputReadLine();

        return received;

        /* Or if cancellation is important to you...
        return Observable.Create<string>(observer =>
            {
                var cancel = Disposable.Create(process.CancelOutputRead);

                return new CompositeDisposable(
                    cancel, 
                    received.Subscribe(observer));
            });
         */
    }
}

找到here。但是当我开始这个过程时

public sealed class ProgramHelper
{
    private readonly Process _program = new Process();
    public IObservable<string> ObservableOutput { get; private set; }

    public ProgramHelper(string programPath, string programArgs)
    {
        _program.StartInfo.FileName = programPath;
        _program.StartInfo.Arguments = programArgs;
    }

    public void StartProgram()
    {
        ConfigService.SaveConfig(
            new Config(
                new Uri(@"http://some.url.com")));

        _program.Start();

        ObservableOutput = _program.StandardOutputObservable();

    }
}

...

[TestFixture]
public class When_program_starts
{
    private ProgramHelper _program;

    [Test]
    public void It_should_not_complain()
    {
       //W
       Action act = () => _program.StartProgram();
       //T
       act.ShouldNotThrow<Exception>();
    }
}

我收到此错误：

“StandardOut尚未重定向或进程尚未开始。”

感谢您的时间。

编辑：将ProgramHelper编辑为

    public ProgramHelper(string programPath, string programArgs)
    {
        _program.StartInfo.FileName = programPath;
        _program.StartInfo.Arguments = programArgs;
        _program.EnableRaisingEvents = true;
        _program.StartInfo.UseShellExecute = false;
        _program.StartInfo.RedirectStandardOutput = true;
    }

但现在它抛出“访问被拒绝的异常”。

我似乎无权以编程方式启动该流程;如果我从控制台启动exe它可以正常工作。

Answer 1

在启动进程后，您正在改变Process.StartInfo属性。

来自Process.StartInfo MSDN documentation：

您可以将StartInfo属性中指定的参数更改为在进程上调用Start方法的时间。启动该过程后，更改StartInfo值不会影响或重新启动关联的进程。

使用反应式扩展观察标准输出

1 个答案: