I am using os.startfile(\'C:\\\\test\\\\sample.exe\')
to launch the application. I don\'t want to know the application’s exit status and I just want to launch t
You should use the subprocess module instead of os.startfile
or os.system
in every case that I'm aware of.
import subprocess
subprocess.Popen([r'C:\test\sample.exe', '-color'])
You could, as @Hackaholic suggests in the comments, do
import os
os.system(r'C:\test\sample.exe -color')
But this is no simpler, and the docs for os recommend the use of subprocess
instead.