我有以下代码:

info = new System.Diagnostics.ProcessStartInfo("TheProgram.exe", String.Join(" ", args));
info.CreateNoWindow = true;
info.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
info.RedirectStandardOutput = true;
info.UseShellExecute = false;
System.Diagnostics.Process p = System.Diagnostics.Process.Start(info);
p.WaitForExit();
Console.WriteLine(p.StandardOutput.ReadToEnd()); //need the StandardOutput contents

我知道我正在启动的进程的输出大约有7MB长。在Windows控制台中运行它可以正常工作。不幸的是,从编程的角度来看,它会无限期地挂在WaitForExit上。还要注意,对于较小的输出(比如3KB),这段代码不会挂起。

ProcessStartInfo中的内部StandardOutput是否可能不能缓冲7MB?如果是,我该怎么办?如果不是,我做错了什么?


当前回答

我读了很多答案,并给出了自己的答案。不确定这个在任何情况下都会修复,但它在我的环境中修复了。我没有使用WaitForExit,而是使用WaitHandle。在输出和错误端信号上等待所有。如果有人能发现可能存在的问题,我会很高兴。或者它是否能帮助别人。对我来说更好,因为不用暂停。

private static int DoProcess(string workingDir, string fileName, string arguments)
{
    int exitCode;
    using (var process = new Process
    {
        StartInfo =
        {
            WorkingDirectory = workingDir,
            WindowStyle = ProcessWindowStyle.Hidden,
            CreateNoWindow = true,
            UseShellExecute = false,
            FileName = fileName,
            Arguments = arguments,
            RedirectStandardError = true,
            RedirectStandardOutput = true
        },
        EnableRaisingEvents = true
    })
    {
        using (var outputWaitHandle = new AutoResetEvent(false))
        using (var errorWaitHandle = new AutoResetEvent(false))
        {
            process.OutputDataReceived += (sender, args) =>
            {
                // ReSharper disable once AccessToDisposedClosure
                if (args.Data != null) Debug.Log(args.Data);
                else outputWaitHandle.Set();
            };
            process.ErrorDataReceived += (sender, args) =>
            {
                // ReSharper disable once AccessToDisposedClosure
                if (args.Data != null) Debug.LogError(args.Data);
                else errorWaitHandle.Set();
            };

            process.Start();
            process.BeginOutputReadLine();
            process.BeginErrorReadLine();

            WaitHandle.WaitAll(new WaitHandle[] { outputWaitHandle, errorWaitHandle });

            exitCode = process.ExitCode;
        }
    }
    return exitCode;
}

其他回答

简介

目前接受的答案不工作(抛出异常),有太多的变通方法,但没有完整的代码。这显然是在浪费很多人的时间,因为这是一个很受欢迎的问题。

结合Mark Byers的回答和Karol Tyl的回答,我根据我想如何使用流程编写了完整的代码。Start方法。

使用

我用它来创建git命令的进度对话框。我是这样使用它的:

    private bool Run(string fullCommand)
    {
        Error = "";
        int timeout = 5000;

        var result = ProcessNoBS.Start(
            filename: @"C:\Program Files\Git\cmd\git.exe",
            arguments: fullCommand,
            timeoutInMs: timeout,
            workingDir: @"C:\test");

        if (result.hasTimedOut)
        {
            Error = String.Format("Timeout ({0} sec)", timeout/1000);
            return false;
        }

        if (result.ExitCode != 0)
        {
            Error = (String.IsNullOrWhiteSpace(result.stderr)) 
                ? result.stdout : result.stderr;
            return false;
        }

        return true;
    }

理论上,您还可以结合stdout和stderr,但我还没有对此进行测试。

Code

public struct ProcessResult
{
    public string stdout;
    public string stderr;
    public bool hasTimedOut;
    private int? exitCode;

    public ProcessResult(bool hasTimedOut = true)
    {
        this.hasTimedOut = hasTimedOut;
        stdout = null;
        stderr = null;
        exitCode = null;
    }

    public int ExitCode
    {
        get 
        {
            if (hasTimedOut)
                throw new InvalidOperationException(
                    "There was no exit code - process has timed out.");

            return (int)exitCode;
        }
        set
        {
            exitCode = value;
        }
    }
}

public class ProcessNoBS
{
    public static ProcessResult Start(string filename, string arguments,
        string workingDir = null, int timeoutInMs = 5000,
        bool combineStdoutAndStderr = false)
    {
        using (AutoResetEvent outputWaitHandle = new AutoResetEvent(false))
        using (AutoResetEvent errorWaitHandle = new AutoResetEvent(false))
        {
            using (var process = new Process())
            {
                var info = new ProcessStartInfo();

                info.CreateNoWindow = true;
                info.FileName = filename;
                info.Arguments = arguments;
                info.UseShellExecute = false;
                info.RedirectStandardOutput = true;
                info.RedirectStandardError = true;

                if (workingDir != null)
                    info.WorkingDirectory = workingDir;

                process.StartInfo = info;

                StringBuilder stdout = new StringBuilder();
                StringBuilder stderr = combineStdoutAndStderr
                    ? stdout : new StringBuilder();

                var result = new ProcessResult();

                try
                {
                    process.OutputDataReceived += (sender, e) =>
                    {
                        if (e.Data == null)
                            outputWaitHandle.Set();
                        else
                            stdout.AppendLine(e.Data);
                    };
                    process.ErrorDataReceived += (sender, e) =>
                    {
                        if (e.Data == null)
                            errorWaitHandle.Set();
                        else
                            stderr.AppendLine(e.Data);
                    };

                    process.Start();

                    process.BeginOutputReadLine();
                    process.BeginErrorReadLine();

                    if (process.WaitForExit(timeoutInMs))
                        result.ExitCode = process.ExitCode;
                    // else process has timed out 
                    // but that's already default ProcessResult

                    result.stdout = stdout.ToString();
                    if (combineStdoutAndStderr)
                        result.stderr = null;
                    else
                        result.stderr = stderr.ToString();

                    return result;
                }
                finally
                {
                    outputWaitHandle.WaitOne(timeoutInMs);
                    errorWaitHandle.WaitOne(timeoutInMs);
                }
            }
        }
    }
}

https://stackoverflow.com/a/17600012/4151626的信用归于EM0

我的应用程序的其他解决方案(包括EM0)仍然处于死锁状态,这是由于内部超时以及派生应用程序同时使用StandardOutput和standderror。以下是对我有效的方法:

Process p = new Process()
{
  StartInfo = new ProcessStartInfo()
  {
    FileName = exe,
    Arguments = args,
    UseShellExecute = false,
    RedirectStandardOutput = true,
    RedirectStandardError = true
  }
};
p.Start();

string cv_error = null;
Thread et = new Thread(() => { cv_error = p.StandardError.ReadToEnd(); });
et.Start();

string cv_out = null;
Thread ot = new Thread(() => { cv_out = p.StandardOutput.ReadToEnd(); });
ot.Start();

p.WaitForExit();
ot.Join();
et.Join();

编辑:添加初始化StartInfo代码示例

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。

我认为使用async,可以有一个更优雅的解决方案,即使同时使用standardOutput和standardError也不会出现死锁:

using (Process process = new Process())
{
    process.StartInfo.FileName = filename;
    process.StartInfo.Arguments = arguments;
    process.StartInfo.UseShellExecute = false;
    process.StartInfo.RedirectStandardOutput = true;
    process.StartInfo.RedirectStandardError = true;

    process.Start();

    var tStandardOutput = process.StandardOutput.ReadToEndAsync();
    var tStandardError = process.StandardError.ReadToEndAsync();

    if (process.WaitForExit(timeout))
    {
        string output = await tStandardOutput;
        string errors = await tStandardError;

        // Process completed. Check process.ExitCode here.
    }
    else
    {
        // Timed out.
    }
}

这是基于Mark Byers的回答。 如果你不在异步方法中,你可以使用string output = tStandardOutput.result;而不是等待

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。)