Passing optional argument in python

Question

I am trying to append a url by passing the parameters by commandline argument. Here is how I am trying:

import argparse
parser = argparse.ArgumentParser(description='Arguments')
parser.add_argument('input', metavar='input', type=str)
parser.add_argument('output', metavar='text', type=str)
args = parser.parse_args()
url = 'https://example.com/?z=12&text='+args.output+'&loc{}'
print url

When I execute

python url.py text.csv hello

It appends the second passed argument to the url. I want to know how to make the second argument optional so that even without providing the second argument I want the url to be printed by concatenating nothing to the url. Here is the output I am expecting:

When both arguments given:

python url.py text.csv hello

The output should be

https://example.com/?z=12&text=hello&loc{}

When single argument given

python url.py text.csv

The output should be

https://example.com/?z=12&text=&loc{}

Setting default? Can you please explain @IgnacioVazquez-Abrams. I am still learning. — csvb
– csvb, Commented Jun 16, 2016 at 11:20

hpaulj · Accepted Answer · 2016-06-16 16:00:18Z

6

Try nargs='?' and define a default:

import argparse
parser = argparse.ArgumentParser(description='Arguments')
parser.add_argument('input', metavar='input')
parser.add_argument('output', metavar='text', nargs='?', default='')
args = parser.parse_args()
url = 'https://example.com/?z=12&text='+args.output+'&loc{}'
print url

When I test it with just one commandline string, args.output is the default:

In [91]: args = parser.parse_args(['one'])

In [92]: args
Out[92]: Namespace(input='one', output='')

In [93]: args = parser.parse_args(['one','two'])

In [94]: args
Out[94]: Namespace(input='one', output='two')

It is best to only use nargs='?' (or '*' or '+') with the last positional argument. It's possible to use it on earlier ones, but there are complications that can send you back with more questions.

edited Jun 16, 2016 at 16:00

answered Jun 16, 2016 at 15:57

hpaulj

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Comments

Community · Accepted Answer · 2017-05-23 11:53:01Z

1

As I see it, you have two options. Either you avoid using argparsealtogether and go for something like:

import sys
args_input  = sys.argv[1]
args_output = sys.argv[2] if len(sys.argv) > 2 else ''
url = 'https://example.com/?z=12&text='+args_output+'&loc{}'
print url

Or you add the - to your optional argument and, as Ignacio's reply suggests, you set a default empty value for the output argument:

import argparse
parser = argparse.ArgumentParser(description='Arguments')
parser.add_argument('input', metavar='input', type=str)
parser.add_argument('-output', metavar='text', type=str, default='')
args = parser.parse_args()
url = 'https://example.com/?z=12&text='+args.output+'&loc{}'
print url

With this second option you'll have to call it like:

python url.py text.csb -output hello

Or

python url.py text.csb

but it is more extensible if you want to add more arguments afterwards.

edited May 23, 2017 at 11:53

CommunityBot

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answered Jun 16, 2016 at 11:39

jaumebonet

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3 Comments

Martin Hallén Over a year ago

If first argument is always passed, it might be a cleaner syntax to only put a - in front of output. Then the program can be called with python url.py text.csb -output hello or python url.py text.csb

jaumebonet Over a year ago

I completely agree! I was just thinking about that!!! I'll re-edit to show your suggestion! Thanks!

csvb Over a year ago

yea. thank you. I opted to add - to the second argument parser. I thought that would be easy addition than changing the rest of the code to sys.argsv[]. I did just like you said to another script using sys.argsv[] @jaumebonet . Thanks again to everyone. :)

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