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

上面的答案没有一个能起作用。

Rob解决方案挂起,“Mark Byers”解决方案获得已处理异常。(我尝试了其他答案的“解决方案”)。

所以我决定提出另一个解决方案:

public void GetProcessOutputWithTimeout(Process process, int timeoutSec, CancellationToken token, out string output, out int exitCode)
{
    string outputLocal = "";  int localExitCode = -1;
    var task = System.Threading.Tasks.Task.Factory.StartNew(() =>
    {
        outputLocal = process.StandardOutput.ReadToEnd();
        process.WaitForExit();
        localExitCode = process.ExitCode;
    }, token);

    if (task.Wait(timeoutSec, token))
    {
        output = outputLocal;
        exitCode = localExitCode;
    }
    else
    {
        exitCode = -1;
        output = "";
    }
}

using (var process = new Process())
{
    process.StartInfo = ...;
    process.Start();
    string outputUnicode; int exitCode;
    GetProcessOutputWithTimeout(process, PROCESS_TIMEOUT, out outputUnicode, out exitCode);
}

这段代码经过调试,工作完美。

其他回答

我也有同样的问题,但原因不同。但是在Windows 8下会发生,而在Windows 7下不会。下面这行似乎是导致这个问题的原因。

pProcess.StartInfo.UseShellExecute = False

解决方案是不禁用UseShellExecute。我现在收到一个Shell弹出窗口,这是不需要的,但比程序等待什么特别的事情发生要好得多。所以我添加了如下的解决方法:

pProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden

现在唯一让我困扰的是为什么在Windows 8下会出现这种情况。

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。

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;
}