Question

我一直在搞乱C＃，在代码的一瞬间，我需要将外部.exe的输出转储到.txt中。我是通过启动cmd.exe然后加载程序及其属性加上>操作符来实现的。但是现在，当我执行程序时，文件甚至都没有创建。同时，如果我在程序中输入传递给cmd的EXACT相同代码：

＆＃34; o：\ steam \ steamapps \ common \反击全球攻势\ bin \ demoinfogo.exe＆＃34; ＆＃34; O：\ Steam \ SteamApps \ common \ Counter-Strike Global Offensive \ csgo \ testfile.dem＆＃34; -gameevents -nofootsteps -deathscsv -nowarmup＆gt; ＆＃34; o：\ steam \ steamapps \ common \反击全球攻势\ decodemp.txt＆＃34;

直接进入命令提示符，确实被转储。我一直在四处寻找，我发现很多的信息，但到目前为止，没有任何帮助我，所以我决定问自己。我附上了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 /?进行访问），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

如果没有其中一个开关，CMD将不会将您提供的字符串解释为执行命令。

当我编写如下的短程序时，它会成功生成一个文件...... 但仅当我使用/C或/K选项时：

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类具有此有用的小属性：

Process.StandardOutput

当Process将文本写入其标准流时，该文本通常显示在控制台上。通过重定向StandardOutput流，您可以操纵或抑制进程的输出。例如，您可以过滤文本，以不同方式对其进行格式化，或将输出写入控制台和指定的日志文件。

您需要做的就是确保将StandardOutput重定向到此流（使用RedirectStandardOutput中的ProcessStartInfo属性），然后您可以读取该流的输出。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process-ish的输出转储到文件中

3 个答案: