从代码C＃运行命令Promt应用程序

Question

This is my code:

var startupPath = Directory.GetParent(Directory.GetCurrentDirectory()).Parent.Parent.FullName; // +\\Common

            System.Diagnostics.Process process = new System.Diagnostics.Process();
            System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
            startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
            startInfo.FileName = "cmd.exe";
            startInfo.Arguments ="/c " + startupPath + @"\Common\sound 1.wav result.wav -tempo=35";
            process.StartInfo = startInfo;
            process.Start();

Pathes are correct. A wanna to start application and set this parameters sound 1.wav result.wav -tempo=35 . What am I doing wrong? content of startInfo.Arguments:

"/c C:\\SOUNDS\\New folder\\TrunscribeHelper\\Common\\sound 1.wav result.wav -tempo=35"

I've tryed to start it directly:

ProcessStartInfo startInfo = new ProcessStartInfo(string.Concat(startupPath + @"\Common", "\\", "sound.exe"));
            startInfo.Arguments = "1.wav result.wav -tempo=35";
            startInfo.UseShellExecute = false;
            System.Diagnostics.Process.Start(startInfo);

But Am I doing wrong now?

Answer 1

Can you try changing this line:

startInfo.Arguments ="/c " + startupPath + @"\Common\sound 1.wav result.wav -tempo=35";

to

startInfo.Arguments ="/c \"" + startupPath + @"\Common\sound"" 1.wav result.wav -tempo=35";

Answer 2

您必须在args中的双引号下设置路径：

"/c \"C:\\SOUNDS\\New folder\\TrunscribeHelper\\Common\\sound 1.wav\" result.wav -tempo=35"