argparse optional subparser (for --version)

Question

I have the following code (using Python 2.7):

# shared command line options, like --version or --verbose
parser_shared = argparse.ArgumentParser(add_help=False)
parser_shared.add_argument('--version', action='store_true')

# the main parser, inherits from `parser_shared`
parser = argparse.ArgumentParser(description='main', parents=[parser_shared])

# several subcommands, which can't inherit from the main parser, since
# it would expect subcommands ad infinitum
subparsers = parser.add_subparsers('db', parents=[parser_shared])

...

args = parser.parse_args()

Now I would like to be able to call this program e.g. with the --version appended to the normal program or some subcommand:

$ prog --version
0.1

$ prog db --version
0.1

Basically, I need to declare optional subparsers. I'm aware that this isn't really supported, but are there any workarounds or alternatives?

Edit: The error message I am getting:

$ prog db --version
# works fine

$ prog --version
usage: ....
prog: error: too few arguments

Answer 1

According to documentation, --version with action='version' (and not with action='store_true') prints automatically the version number:

parser.add_argument('--version', action='version', version='%(prog)s 2.0')

Answer 2

FWIW, I ran into this also, and ended up "solving" it by not using subparsers (I already had my own system for printing help, so didn't lose anything there).

Instead, I do this:

parser.add_argument("command", nargs="?",
                    help="name of command to execute")

args, subcommand_args = parser.parse_known_args()

...and then the subcommand creates its own parser (similar to a subparser) which operates only on subcommand_args.