在python中传递可选参数

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{}

Answer 1

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.

Answer 2

As I see it, you have two options. Either you avoid using argparse altogether 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.