Python argv validation

Question

I need to check if the user is giving the input file and the name of the output, and I'm doing the following:

def main():
    if len(argv) > 2:
        script, file_in, file_out = argv
        execute_code(file_in, file_out)
    else:
        print "Wrong number of arguments!"
        print "Usage: python script.py filename_input filename_output"


if __name__ == '__main__':
    main()

Is there other way to check if the argv arguments are correct?

Answer 1

You'd use argparse:

The argparse module makes it easy to write user-friendly command-line interfaces. The program defines what arguments it requires, and argparse will figure out how to parse those out of sys.argv.

For example your main function could be rewritten as

import argparse

def main():
    parser = argparse.ArgumentParser()
    parser.add_argument('file_in', help='input file')
    parser.add_argument('file_out', help='output file')
    args = parser.parse_args()
    execute_code(args.file_in, args.file_out)

if __name__ == '__main__':
    main()

argparse will perform validation for you and display fairly helpful error messages if some of the required arguments are missing:

vaultah@base:~$ python3 /home/vaultah/untitled.py
usage: untitled.py [-h] file_in file_out
untitled.py: error: the following arguments are required: file_in, file_out
vaultah@base:~$ python3 /home/vaultah/untitled.py in
usage: untitled.py [-h] file_in file_out
untitled.py: error: the following arguments are required: file_out

Additionally, it will generate a help message

vaultah@base:~$ python3 /home/vaultah/untitled.py -h
usage: untitled.py [-h] file_in file_out

positional arguments:
  file_in     input file
  file_out    output file

optional arguments:
  -h, --help  show this help message and exit