how to add multiple argument options in python usi

Question

My Requirement:

For now when I run my python application with this command

python main.py -d listhere/users.txt

The program will run and save the result file as predefined name say reports.txt

Now I want to add this functionality to allow users to choose what to put the filename and where to save as so

python main.py -d -o output/newfilname -i listhere/users.txt

Everything is same but I want another argument -o to be passed which will determine the filpath and name to be saved. How do I do it. What is the best way to handle or combine multiple options.

I tried this

    parser = argparse.ArgumentParser(description = "CHECK-ACCESS REPORTING.")
    parser.add_argument('--user','-d', nargs='?')
    parser.add_argument('--output','-d -o', nargs='?')
    parser.add_argument('--input','-i', nargs='?')
    args = parser.parse_args(sys.argv[1:])

   if args.output and args.input:
        #operation that involves output filename too
   elif args.user and not args.input:
       #default operation only
   else:
      #notset

I am getting this error when trying to solve the issue this way

Error:

report.py: error: unrecognized arguments: -o listhere/users.txt

Answer 1

Do you want to have something like this:

import argparse

def main():
    parser = argparse.ArgumentParser(
        description='Check-Access Reporting.',
        formatter_class=argparse.ArgumentDefaultsHelpFormatter,
    )
    parser.add_argument(
        '-d',
        dest='discrepancy',
        action='store_true',
        help='Generate discrepancy report.',
    )
    parser.add_argument(
        '--input',
        '-i',
        default='users.txt',
        help='Input file for the report.',
    )
    parser.add_argument(
        '--output',
        '-o',
        default='reports.txt',
        help='Output file for the report.',
    )
    args = parser.parse_args()

    if args.discrepancy:
        print('Report type: {}'.format(args.report_type))
        print('Input file: {}'.format(args.input))
        print('Output file: {}'.format(args.output))
    else:
        print('Report type is not specified.')

if __name__ == '__main__':
    main()

Result of option --help:

usage: ptest_047.py [-h] [-d] [--input INPUT] [--output OUTPUT]

Check-Access Reporting.

optional arguments:
  -h, --help            show this help message and exit
  -d                    generate discrepancy report (default: False)
  --input INPUT, -i INPUT
                        input file for the report (default: users.txt)
  --output OUTPUT, -o OUTPUT
                        output file for the report (default: reports.txt)

Without any option (or missing option -d):

Report type is not specified.

With option -d:

Report type: discrepancy
Input file: users.txt
Output file: reports.txt

With -d --input input.txt --output output.txt:

Report type: discrepancy
Input file: input.txt
Output file: output.txt

Answer 2

A nargs='?' flagged option works in 3 ways

parser.add_argument('-d', nargs='?', default='DEF', const='CONST')

commandline:

foo.py -d value # => args.d == 'value'
foo.py -d       # => args.d == 'CONST'
foo.py          # => args.d == 'DEF'

https://docs.python.org/3/library/argparse.html#const

Taking advantage of that, you shouldn't need anything like this erroneous -d -o flag.

If you don't use the const parameter, don't use '?'

parser.add_argument('--user','-u', nargs='?', const='CONST', default='default_user')
parser.add_argument('--output','-o', default='default_outfile')
parser.add_argument('--input','-i', default='default_infile')