我知道这是晚餐，但是，在读完这一整页后，没有一个解决方案对我来说是有效的，尽管我没有尝试Muhammad Rehan，因为代码有点难理解，尽管我猜他是在正确的轨道上。当我说它不工作时，这并不完全正确，有时它会工作得很好，我猜这与EOF标记之前的输出长度有关。

不管怎样，对我来说有效的解决方案是使用不同的线程来读取StandardOutput和StandardError并写入消息。

        StreamWriter sw = null;
        var queue = new ConcurrentQueue<string>();

        var flushTask = new System.Timers.Timer(50);
        flushTask.Elapsed += (s, e) =>
        {
            while (!queue.IsEmpty)
            {
                string line = null;
                if (queue.TryDequeue(out line))
                    sw.WriteLine(line);
            }
            sw.FlushAsync();
        };
        flushTask.Start();

        using (var process = new Process())
        {
            try
            {
                process.StartInfo.FileName = @"...";
                process.StartInfo.Arguments = $"...";
                process.StartInfo.UseShellExecute = false;
                process.StartInfo.RedirectStandardOutput = true;
                process.StartInfo.RedirectStandardError = true;

                process.Start();

                var outputRead = Task.Run(() =>
                {
                    while (!process.StandardOutput.EndOfStream)
                    {
                        queue.Enqueue(process.StandardOutput.ReadLine());
                    }
                });

                var errorRead = Task.Run(() =>
                {
                    while (!process.StandardError.EndOfStream)
                    {
                        queue.Enqueue(process.StandardError.ReadLine());
                    }
                });

                var timeout = new TimeSpan(hours: 0, minutes: 10, seconds: 0);

                if (Task.WaitAll(new[] { outputRead, errorRead }, timeout) &&
                    process.WaitForExit((int)timeout.TotalMilliseconds))
                {
                    if (process.ExitCode != 0)
                    {
                        throw new Exception($"Failed run... blah blah");
                    }
                }
                else
                {
                    throw new Exception($"process timed out after waiting {timeout}");
                }
            }
            catch (Exception e)
            {
                throw new Exception($"Failed to succesfully run the process.....", e);
            }
        }
    }

希望这能帮助那些认为这很难的人!

我认为这是一个简单和更好的方法(我们不需要AutoResetEvent):

public static string GGSCIShell(string Path, string Command)
{
    using (Process process = new Process())
    {
        process.StartInfo.WorkingDirectory = Path;
        process.StartInfo.FileName = Path + @"\ggsci.exe";
        process.StartInfo.CreateNoWindow = true;
        process.StartInfo.RedirectStandardOutput = true;
        process.StartInfo.RedirectStandardInput = true;
        process.StartInfo.UseShellExecute = false;

        StringBuilder output = new StringBuilder();
        process.OutputDataReceived += (sender, e) =>
        {
            if (e.Data != null)
            {
                output.AppendLine(e.Data);
            }
        };

        process.Start();
        process.StandardInput.WriteLine(Command);
        process.BeginOutputReadLine();


        int timeoutParts = 10;
        int timeoutPart = (int)TIMEOUT / timeoutParts;
        do
        {
            Thread.Sleep(500);//sometimes halv scond is enough to empty output buff (therefore "exit" will be accepted without "timeoutPart" waiting)
            process.StandardInput.WriteLine("exit");
            timeoutParts--;
        }
        while (!process.WaitForExit(timeoutPart) && timeoutParts > 0);

        if (timeoutParts <= 0)
        {
            output.AppendLine("------ GGSCIShell TIMEOUT: " + TIMEOUT + "ms ------");
        }

        string result = output.ToString();
        return result;
    }
}

Mark Byers的回答非常棒，但我想补充以下内容:

OutputDataReceived和ErrorDataReceived委托需要在outputWaitHandle和errorWaitHandle被释放之前被移除。如果进程在超过超时时间后继续输出数据，然后终止，则outputWaitHandle和errorWaitHandle变量将在被释放后被访问。

(仅供参考，我不得不加上这个警告作为回答，因为我不能评论他的帖子。)

我试图通过考虑Mark Byers, Rob, stevejay的回答，创建一个类来解决你使用异步流读取的问题。这样做，我意识到有一个与异步流程输出流读取相关的错误。

我在微软报告了这个漏洞:https://connect.microsoft.com/VisualStudio/feedback/details/3119134

简介:

You can't do that: process.BeginOutputReadLine(); process.Start(); You will receive System.InvalidOperationException : StandardOut has not been redirected or the process hasn't started yet. ============================================================================================================================ Then you have to start asynchronous output read after the process is started: process.Start(); process.BeginOutputReadLine(); Doing so, make a race condition because the output stream can receive data before you set it to asynchronous:

process.Start(); 
// Here the operating system could give the cpu to another thread.  
// For example, the newly created thread (Process) and it could start writing to the output
// immediately before next line would execute. 
// That create a race condition.
process.BeginOutputReadLine();

============================================================================================================================ Then some people could say that you just have to read the stream before you set it to asynchronous. But the same problem occurs. There will be a race condition between the synchronous read and set the stream into asynchronous mode. ============================================================================================================================ There is no way to acheive safe asynchronous read of an output stream of a process in the actual way "Process" and "ProcessStartInfo" has been designed.

对于您的情况，您最好使用其他用户建议的异步读取。但是您应该意识到，由于竞态条件，您可能会错过一些信息。

这篇文章可能过时了，但我发现它通常挂起的主要原因是由于堆栈溢出的redirectStandardoutput或如果你有redirectstandderror。

由于输出数据或错误数据比较大，在不确定的时间内仍在处理，会造成挂起时间。

为了解决这个问题:

p.StartInfo.RedirectStandardoutput = False
p.StartInfo.RedirectStandarderror = False

我最终使用的解决方案来避免所有的复杂性:

var outputFile = Path.GetTempFileName();
info = new System.Diagnostics.ProcessStartInfo("TheProgram.exe", String.Join(" ", args) + " > " + outputFile + " 2>&1");
info.CreateNoWindow = true;
info.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
info.UseShellExecute = false;
System.Diagnostics.Process p = System.Diagnostics.Process.Start(info);
p.WaitForExit();
Console.WriteLine(File.ReadAllText(outputFile)); //need the StandardOutput contents

所以我创建了一个临时文件，通过使用> outputfile > 2>&1将输出和错误重定向到它，然后在进程完成后读取文件。

其他解决方案适用于希望对输出执行其他操作的场景，但对于简单的操作，这可以避免很多复杂性。