Take string argument from command line?

Question

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?

dm03514 · Accepted Answer · 2012-10-03 18:22:40Z

9

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

answered Oct 3, 2012 at 18:22

dm03514

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Junuxx · Accepted Answer · 2012-10-03 18:22:29Z

8

Check sys.argv. It gives a list with the name of the script and any command line arguments.

Example script:

import sys
print sys.argv

Calling it:

> python script.py foobar baz
['script.py', 'foobar', 'baz']

answered Oct 3, 2012 at 18:22

Junuxx

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Damian Schenkelman · Accepted Answer · 2012-10-03 18:22:34Z

4

If you are looking for the first parameter sys.argv[1] does the trick. More info here.

answered Oct 3, 2012 at 18:22

Damian Schenkelman

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nishnet2002 · Accepted Answer · 2022-02-23 20:31:52Z

2

Try argparse's default parameter, its well documented.

import argparse

parser = argparse.ArgumentParser(description='Create Configuration')
parser.add_argument('--file-name', type=str, help='optional filename', 
       default="defaultFile.json")

args = parser.parse_args()

print(args.file_name)

Output:

$ python open.py --file-name option1
option1

$ python open.py 
defaultFile.json

Alternative library:

click library for arg parsing.

answered Feb 23, 2022 at 20:31

nishnet2002

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