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 the exe.
I need to pass the argument to that exe like 'C:\\test\\sample.exe' -color
Please suggest a method to run this in Python.
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.
os.startfile opens in a new process. I've edited my answer; you should use subprocess.Popen if you don't want the call to block.'C:\\foo' OR raw strings r'C:\foo'. Not both. If you use both, you will have too many backslashes in your file paths. That is probably why the os.system call in your comment is failing.[r'C:\data\ANSYS.exe', '-b', '-i', 'cdb_wb.txt']. If this doesn't work for you, please ask a new question.Create a batch file sam_ple.bat with the following commands and arguments
cd C:\test\
start sample.exe -color
Then put sam_ple.bat in the same directory as your script.py file
Enter the following line of code in python to launch the exe:
os.startfile('.\sam_ple.bat')
os.system("command")