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Home/ Questions/Q 8616231
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T05:34:08+00:00 2026-06-12T05:34:08+00:00

I need to take an optional argument when running my Python script: python3 myprogram.py

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I need to take an optional argument when running my Python script:

python3 myprogram.py afile.json

or

python3 myprogram.py

This is what I’ve been trying:

filename = 0
parser = argparse.ArgumentParser(description='Create Configuration')
parser.add_argument('filename', type=str,
                   help='optional filename')

if filename is not 0:
    json_data = open(filename).read()
else:
    json_data = open('defaultFile.json').read()

I was hoping to have the filename stored in my variable called “filename” if it was provided. Obviously this isn’t working. Any advice?

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  1. Editorial Team
    Editorial Team
    2026-06-12T05:34:09+00:00Added an answer on June 12, 2026 at 5:34 am

    Please read the tutorial carefully. http://docs.python.org/howto/argparse.html

    i believe you need to actually parse the arguments:

    parser = argparse.ArgumentParser()
    args = parser.parse_args()
    

    then filename will be come available args.filename

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