Argparse: Don't Show Usage On -h

The code from argparse import ArgumentParser p = ArgumentParser(description = 'foo') p.add_argument('-b', '--bar', help = 'a description') p.parse_args() ...results in the output:

Solution 1:

definitely possible -- but I'm not sure about documented ...

from argparse import ArgumentParser,SUPPRESS
p = ArgumentParser(description = 'foo',usage=SUPPRESS)
p.add_argument('-b', '--bar', help = 'a description')
p.parse_args()

---

From reading the source, I've hacked a little something together which seems to work when displaying error messages as well ... warning -- This stuff is mostly undocumented, and therefore liable to change at any time :-)

from argparse import ArgumentParser,SUPPRESS
import sys as _sys
from gettext import gettext as _

classMyParser(ArgumentParser):
    deferror(self, message):    
        usage = self.usage
        self.usage = None
        self.print_usage(_sys.stderr)
        self.exit(2, _('%s: error: %s\n') % (self.prog, message))
        self.usage = usage


p = MyParser(description = 'foo',usage=SUPPRESS)
p.add_argument('-b', '--bar', help = 'a description')
p.parse_args()

Solution 2:

Note: to not show usage for a specific argument, use

parser.add_argument('--foo', help=argparse.SUPPRESS)

per the documentation.

Solution 3:

One additional note: if you don't want help at all you can construct the parser with add_help=False

def parse_args():
    parser = argparse.ArgumentParser(add_help=False)
    parser.add_argument("arg")
    return parser.parse_known_args()

def main():
    args, remaining = parse_args()
    print(args)
    print(remaining)

if __name__ == '__main__':
    main()


user@host % python3 /tmp/test.py  f
Namespace(arg='f')
[]
user@host % python3 /tmp/test.py  f -h
Namespace(arg='f')
['-h']
user@host % python3 /tmp/test.py  f -h --help
Namespace(arg='f')
['-h', '--help']
user@host %