C# - 我无法将我的进程输出转储到文件中

Question

我一直在摆弄 C#，在代码的某个时刻，我需要将外部 .exe 的输出转储到 .txt 中。我通过启动 cmd.exe 然后加载程序及其属性以及 > 运算符来实现。但是现在，当我执行程序时，甚至都没有创建文件。同时，如果我在程序中输入传递给 cmd 的完全相同的代码：

"o:\steam\steamapps\common\counter-strike global offensive\bin\demoinfogo.exe" "O:\Steam\SteamApps\common\Counter-Strike Global Offensive\csgo\testfile.dem" -gameevents -nofootsteps -deathscsv -nowarmup > "o:\steam\steamapps\common\counter-strike global offensive\demodump.txt"

直接进入命令提示符，它确实被转储。我一直在四处寻找，发现了很多的信息，但遗憾的是到目前为止没有任何帮助，所以我决定问问自己。我附上我认为与此相关的代码块。

ProcessStartInfo startInfo = new ProcessStartInfo();

startInfo.CreateNoWindow = false;
startInfo.UseShellExecute = true;
startInfo.FileName = "CMD.exe";
startInfo.WindowStyle = ProcessWindowStyle.Hidden;

if (checkBox1.Checked)
{
    arguments += " -gameevents";
    if (checkBox2.Checked)
    {
        arguments += " -nofootsteps";
    }
    if (checkBox3.Checked)
    {
        arguments += " -extrainfo";
    }
}
if (checkBox4.Checked)
{
    arguments += " -deathscsv";
    if (checkBox5.Checked)
    {
        arguments += " -nowarmup";
    }
}

if (checkBox6.Checked)
{
    arguments += " -stringtables";
}
if (checkBox7.Checked)
{
    arguments += " -datatables";
}
if (checkBox8.Checked)
{
    arguments += " -packetentites";
}
if (checkBox9.Checked)
{
    arguments += " -netmessages";
}
if (dumpfilepath == string.Empty)
{
    dumpfilepath =  getCSGOInstallationPath() + @"\demodump.txt";
}

baseOptions = @"""" + demoinfogopath + @"""" + " " + @"""" + demofilepath + @"""" + arguments;
startInfo.Arguments = baseOptions + " > " + @"""" + dumpfilepath + @"""";

try  
{
    using (exeProcess = Process.Start(startInfo))
         ....a bunch of code...

Answer 1

如果您查看 CMD 的帮助（通过键入 CMD /? 访问），您将看到以下选项：

/C   Carries out the command specified by string and then terminates 
/K   Carries out the command specified by string but remains

如果没有这些开关之一，CMD 将不会将您提供的字符串解释为要执行的命令。

当我编写如下所示的短程序时，它成功生成了一个文件...但前提是我使用 /C 或 /K选项：

ProcessStartInfo startInfo = new ProcessStartInfo();

startInfo.CreateNoWindow = false;
startInfo.UseShellExecute = true;
startInfo.FileName = "CMD.exe";
startInfo.WindowStyle = ProcessWindowStyle.Hidden;

var command = @"echo test > c:\users\myusername\Desktop\test.txt";
var args = "/C " + command;
startInfo.Arguments = args;

using (var process = Process.Start(startInfo)) { }

Answer 2

您正在创建的 Process class 有这个有用的小东西属性:

Process.StandardOutput

When a Process writes text to its standard stream, that text is normally displayed on the console. By redirecting the StandardOutput stream, you can manipulate or suppress the output of a process. For example, you can filter the text, format it differently, or write the output to both the console and a designated log file.

您需要做的就是确保将 StandardOutput 重定向到此流（使用 ProcessStartInfo 中的 RedirectStandardOutput 属性），然后您可以从中读取输出那条溪流。这是 MSDN 示例代码，略有删节：

Process myProcess = new Process();
ProcessStartInfo myProcessStartInfo = new ProcessStartInfo(args[0], "spawn");
myProcessStartInfo.UseShellExecute = false; // important!
myProcessStartInfo.RedirectStandardOutput = true; // also important!
myProcess.StartInfo = myProcessStartInfo;
myProcess.Start();

// Here we're reading the process output's first line:

StreamReader myStreamReader = myProcess.StandardOutput;
string myString = myStreamReader.ReadLine();
Console.WriteLine(myString);

Answer 3

//Hi you could try this to build your process like this.
public class Launcher
{
    public Process CurrentProcess;
    public string result = null;

    public Process Start()
    {
        CurrentProcess = new Process
        {
            StartInfo =
            {
                UseShellExecute = false,
                CreateNoWindow = true,
                RedirectStandardOutput = true,
                RedirectStandardError = true,
                RedirectStandardInput = true,
                WorkingDirectory = @"C:\",
                FileName = Path.Combine(Environment.SystemDirectory, "cmd.exe")
            }
        };
        CurrentProcess.Start();

        return CurrentProcess;
    }

    //Start the process to get the output you want to add to your .txt file:
    private void writeOuput()
    {
        Currentprocess = new process();
        Start()

        CurrentProcess.StandardInput.WriteLine("Your CMD");
        CurrentProcess.StandardInput.Close();

        result = CurrentProcess.StandardOutput.ReadLine();
        CurrentProcess.StandardOutput.Close()

        //Then to put the result in a .txt file:
        System.IO.File.WriteAllText (@"C:\path.txt", result);
    }
}

}

C# - 我无法将我的进程输出转储到文件中

C# - I can't dump to a file the output of my process-ish

c#

variables

dump

process