相当于Process.Start(),没有单独的参数
本文关键字:没有单 参数 Process Start 相当于 | 更新日期: 2023-09-27 17:53:12
我正在编写一个需要运行任意命令的简单应用程序,例如:
powershell -File myscript.ps1
cmd /C "ping localhost"
Process.Start()将是完美的,除了它需要将参数作为单独的参数给出。最初我认为我可以在第一个空格字符上分割字符串,但是如果可执行路径被引用并包含空格怎么办?是否有像Process.Start()这样的东西允许您只给它一个字符串,带或不带参数,并让它像粘贴到命令提示符一样执行它?
为什么不直接通过cmd/C运行所有内容呢?
Process.Start("cmd", "/C " + command);
直接调用CreateProcess
函数:
public static class Native
{
[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Unicode)]
private struct STARTUPINFO
{
public Int32 cb;
public string lpReserved;
public string lpDesktop;
public string lpTitle;
public Int32 dwX;
public Int32 dwY;
public Int32 dwXSize;
public Int32 dwYSize;
public Int32 dwXCountChars;
public Int32 dwYCountChars;
public Int32 dwFillAttribute;
public Int32 dwFlags;
public Int16 wShowWindow;
public Int16 cbReserved2;
public IntPtr lpReserved2;
public IntPtr hStdInput;
public IntPtr hStdOutput;
public IntPtr hStdError;
}
[StructLayout(LayoutKind.Sequential)]
private struct PROCESS_INFORMATION
{
public IntPtr hProcess;
public IntPtr hThread;
public int dwProcessId;
public int dwThreadId;
}
[DllImport("kernel32.dll", SetLastError = true, CharSet = CharSet.Auto)]
private static extern bool CreateProcess(
string lpApplicationName,
string lpCommandLine,
IntPtr lpProcessAttributes,
IntPtr lpThreadAttributes,
bool bInheritHandles,
uint dwCreationFlags,
IntPtr lpEnvironment,
string lpCurrentDirectory,
[In] ref STARTUPINFO lpStartupInfo,
out PROCESS_INFORMATION lpProcessInformation);
public static void CreateProcessFromCommandLine(string commandLine)
{
var si = new STARTUPINFO();
var pi = new PROCESS_INFORMATION();
CreateProcess(
null,
commandLine,
IntPtr.Zero,
IntPtr.Zero,
false,
0,
IntPtr.Zero,
null,
ref si,
out pi);
}
}
internal class Program
{
public static void Main()
{
Native.CreateProcessFromCommandLine("ping google.com -t");
Console.Read();
}
}
您可以使用Path
类来计算命令行字符串,并像这样运行Process.Start
:
// This is our command string
var cmd = @"C:'Program Files'Sub folder 1'executable name.exe -arg1 -arg2 -argN";
var exeName = Path.GetFileName(cmd); // "executable name.exe -arg1 -arg2 -argN"
var borderPos = exeName.IndexOf(".exe") // or .cmd, .bat, etc.
borderPos = borderPos == -1 ? exeName.IndexOf(" ") : borderPos + 4;
var arguments = exeName.Substring(borderPos).Trim();
var program = cmd.Substring(0, cmd.Length - arguments.Length);
Process.Start(program, aguments);
类似这样的代码应该可以做到这一点,但是上面的代码是未经测试的…