重定向子进程的输出（stdout，标准错误）在Visual Studio Output窗口 [英] Redirect the output (stdout, stderr) of a child process to the Output window in Visual Studio

问题描述

目前，我开始从我的C＃程序用一个批处理文件：

  System.Diagnostics.Process.Start（@DoSomeStuff.bat）;
 

我想能够做的是，子进程的输出（stdout和stderr）重定向到Visual Studio中的输出窗口（具体的Visual C＃防爆preSS 2008）。

有没有办法做到这一点？

（另外：这样，它不是所有缓冲了再吐出来，当子进程结束Output窗口）

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（BTW：现在我可以得到标准输出（而不是标准错误）的的父的过程出现在输出窗口，通过我的节目一个Windows应用程序，而不是控制台应用程序。这打破，如果该程序的Visual Studio之外运行，但这是我的具体情况确定。）

解决方案

  process.StartInfo.CreateNoWindow = TRUE;
process.StartInfo.UseShellExecute = FALSE;
process.StartInfo.RedirectStandardOutput =真;
process.OutputDataReceived + =（发件人，参数）=＆GT; Console.WriteLine（args.Data）;
process.BeginOutputReadLine（）;
 

为错误

同样的想法，只需更换输出这些方法/属性名。

At the moment I am starting a batch file from my C# program with:

System.Diagnostics.Process.Start(@"DoSomeStuff.bat");

What I would like to be able to do is redirect the output (stdout and stderr) of that child process to the Output window in Visual Studio (specifically Visual C# Express 2008).

Is there a way to do that?

(Additionally: such that it's not all buffered up and then spat out to the Output window when the child process finishes.)

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(BTW: At the moment I can get stdout (but not stderr) of the parent process to appear in the Output window, by making my program a "Windows Application" instead of a "Console Application". This breaks if the program is run outside Visual Studio, but this is ok in my particular case.)

解决方案

process.StartInfo.CreateNoWindow = true;
process.StartInfo.UseShellExecute = false;
process.StartInfo.RedirectStandardOutput = true;
process.OutputDataReceived  += (sender, args) => Console.WriteLine(args.Data);
process.BeginOutputReadLine();

Same idea for Error, just replace Output in those method/property names.

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