python argparse: arg with no flag
argparse, python, python-3.x
Solution
Yes.
You have multiple solutions:
- As Mrav mentioned, you can use the system argument (sys.argv[0...])
Or use argparse. From the documentation (which is python3 compliant), you can do this way:
if __name__ == '__main__':
parser = ArgumentParser()
parser.add_argument('file')
parser.add_argument('revs', metavar='N', type=int, nargs='+', help='revisions')
res = parser.parse_args()
pprint(res)
And you can see the result:
$ ./test.py settings.json 1 2 3
Namespace(file='settings.json', revs=[1, 2, 3])
Problem
i've the following code: ``` parser.add_argument('file', help='file to test') parser.add_argument('-revs', help='range of versions', nargs='+', default=False) ``` Is there a way to not use the flag `-revs` when use it, like this: ``` ./somescript.py settings.json 1 2 3 4 ```