I want to use argparse to only show help and usage and nothing else.

My current code looks like:

import argparse
parser = argparse.ArgumentParser(prog='remove-duplicate-lines-except-blank-lines',
                                 usage='%(prog)s [options] [files...]',
                                 description='Remove duplicate line except white space')
args = parser.parse_args()

To get help I can run:

$ python origname.py -h

and

$ python origname.py --help

So far, so good. However, when I try to use file names without options, for example:

$ python origname.py inputfile.txt optputfile.txt

If gives me usage and an error.

usage: origname [options] [files...]
origname: error: unrecognized arguments: inputfile.txt optputfile.txt

How can I pass arguments without options when using argparse?

Just to be clear, I am handling the rest of the argument (ex. inputfile.txt optputfile.txt) etc. manually using sys.argv so I do not need argparse for that.

CodePudding user response：

Instead of using parser.parse_args() use parser.parse_known_args().

import argparse
parser = argparse.ArgumentParser(prog='remove-duplicate-lines-except-blank-lines',
                                 usage='%(prog)s [options] [files...]',
                                 description='Remove duplicate line except white space')
parser.parse_known_args()

==>python foo.py --help
usage: remove-duplicate-lines-except-blank-lines [options] [files...]

Remove duplicate line except white space

optional arguments:
  -h, --help  show this help message and exit


==>python foo.py foo.txt
# nothing happens here