Run Python script via CMD

Question

I have the following file: up.py

in this file:

def main(a_param, b_param, c_param):
    // Code
if __name__ == '__main__':
    exit(main())

I want to run this python file via the CMD, so I write this line:

python up.py False True False

But I get the next error:

TypeError: main() takes exactly 3 arguments (0 given)

Answer 1

This has nothing to do with CMD. Your main function expects three arguments, but you aren't passing any; you call it directly from your if __name__ == '__main__' block with just main() .

Either get the arguments (eg from sys.argv ) within that block and pass them to main, or remove the arguments from the function signature and get them within main.

Answer 2

You are trying to call your main function without arguments event though it requires 3 ( a_param , b_param and c_param ).

The command line parameters are stored in sys.argv . To call the main function with the first 3 command line parameters, you could this:

import sys

if __name__ == '__main__':
    main(*sys.argv[1:4])

To clarify, * unpacks the argument list so main(*sys.argv[1:4]) is equivalent to main(sys.argv[1], sys.argv[2], sys.argv[3])

Answer 3

This code works for me

def main(a_param, b_param, c_param):
    # Code
    if __name__ == '__main__':
        exit(main())

then:

$ python up.py False True False