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;
}
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
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。