Prevent child process from showing a shell window in c#

c#, process, processstartinfo, windows

Solution

You should try to use

ProcessStartInfo si = new ProcessStartInfo();
si.WindowStyle = ProcessWindowStyle.Hidden;
si.CreateNoWindow = true;
si.UseShellExecute = false;

MSDN refs

Problem

I'm using ffmpeg to compile videos, and I'd like to prevent it from displaying a console when performing actions. Here's how I start ffmpeg: ``` ProcessStartInfo si = new ProcessStartInfo(); si.Arguments = string.Format("-y -loop 1 -t " + DucationToString(frameDuration) + " -r 25 -f image2 -i \"{0}\" \"{1}\"", item.Value, otpt); si.FileName = "ffmpeg"; si.UseShellExecute = false; Process.Start(si).WaitForExit(); ``` No matter the settings I try in `ProcessStartInfo`, the console always shows up. How do I prevent the console from being shown when creating child process?

Original source