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?
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。
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);
}
}
}
}
}
问题是，如果你重定向StandardOutput和/或StandardError，内部缓冲区可能会满。不管你用什么顺序，都会有一个问题:
如果在读取StandardOutput之前等待进程退出，进程可能会阻塞尝试写入它，因此进程永远不会结束。
如果你使用ReadToEnd从StandardOutput中读取，那么如果进程从未关闭StandardOutput(例如，如果它从未终止，或者如果它被阻塞写入到standderror)，那么你的进程就会阻塞。
解决方案是使用异步读取来确保缓冲区不会被填满。为了避免死锁并从StandardOutput和standderror收集所有输出，你可以这样做:
编辑:请参阅下面的回答，了解如何在超时发生时避免ObjectDisposedException。
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;
StringBuilder output = new StringBuilder();
StringBuilder error = new StringBuilder();
using (AutoResetEvent outputWaitHandle = new AutoResetEvent(false))
using (AutoResetEvent errorWaitHandle = new AutoResetEvent(false))
{
process.OutputDataReceived += (sender, e) => {
if (e.Data == null)
{
outputWaitHandle.Set();
}
else
{
output.AppendLine(e.Data);
}
};
process.ErrorDataReceived += (sender, e) =>
{
if (e.Data == null)
{
errorWaitHandle.Set();
}
else
{
error.AppendLine(e.Data);
}
};
process.Start();
process.BeginOutputReadLine();
process.BeginErrorReadLine();
if (process.WaitForExit(timeout) &&
outputWaitHandle.WaitOne(timeout) &&
errorWaitHandle.WaitOne(timeout))
{
// Process completed. Check process.ExitCode here.
}
else
{
// Timed out.
}
}
}
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。
在阅读了这里所有的帖子之后，我确定了Marko avlijaovic的统一解决方案。
然而，它并没有解决我所有的问题。
在我们的环境中，我们有一个Windows服务，计划运行数百个不同的.bat .cmd .exe，…等等，这些文件积累了多年，由许多不同的人以不同的风格编写。我们无法控制程序和脚本的编写，我们只负责调度、运行和成功/失败的报告。
所以我尝试了这里所有的建议，并取得了不同程度的成功。Marko的回答几乎是完美的，但是当作为服务运行时，它并不总是捕获标准输出。我一直没搞清楚为什么不行。
我们发现在所有情况下都有效的唯一解决方案是:http://csharptest.net/319/using-the-processrunner-class/index.html
