python argparse: arg with no flag

Question

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

FlogFR · Accepted Answer · 2014-05-21 08:27:59Z

38

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])

answered May 21, 2014 at 8:27

FlogFR

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