Pass arguments to executable in Python
Question:
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.
Answers:
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:testsample.exe', '-color'])
You could, as @Hackaholic suggests in the comments, do
import os
os.system(r'C:testsample.exe -color')
But this is no simpler, and the docs for os recommend the use of subprocess instead.
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')
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:testsample.exe', '-color'])
You could, as @Hackaholic suggests in the comments, do
import os
os.system(r'C:testsample.exe -color')
But this is no simpler, and the docs for os recommend the use of subprocess instead.
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')