在使用argparse和'nargs =*'参数声明所需参数时断言

Question

以下是解析参数的当前代码:

parser = argparse.ArgumentParser()

parser.add_argument('-a', '--action', required=True, metavar='', nargs=1, help='{block|release|clear|show|show_extended|}')
parser.add_argument('-i', '--interface', required=True, metavar='', nargs=1, help='interface name')
parser.add_argument('-d', '--debug', action='store_true', help='debug prints')
parser.add_argument('--ips', metavar='ips', nargs='*', help='ip addresses to block')
parser.add_argument('--handles', metavar='handles', nargs='*', help='filters handles to delete (usually 800::xxx), run with [show] to see')
args = parser.parse_args()

当我尝试执行它而不键入-a或-i时,我得到一个断言:

Traceback (most recent call last):
  File "./block_traffic.py", line 112, in <module>
    args = parser.parse_args()
  File "/usr/lib/python2.7/argparse.py", line 1690, in parse_args
    args, argv = self.parse_known_args(args, namespace)
  File "/usr/lib/python2.7/argparse.py", line 1722, in parse_known_args
    namespace, args = self._parse_known_args(args, namespace)
  File "/usr/lib/python2.7/argparse.py", line 1946, in _parse_known_args
    self.error(_('argument %s is required') % name)
  File "/usr/lib/python2.7/argparse.py", line 2362, in error
    self.print_usage(_sys.stderr)
  File "/usr/lib/python2.7/argparse.py", line 2324, in print_usage
    self._print_message(self.format_usage(), file)
  File "/usr/lib/python2.7/argparse.py", line 2280, in format_usage
    return formatter.format_help()
  File "/usr/lib/python2.7/argparse.py", line 281, in format_help
    help = self._root_section.format_help()
  File "/usr/lib/python2.7/argparse.py", line 211, in format_help
    func(*args)
  File "/usr/lib/python2.7/argparse.py", line 332, in _format_usage
    assert ' '.join(opt_parts) == opt_usage
AssertionError

我注意到断言异常消失了,因为我删除了nargs='*'参数(两个中的一个)

我想要的是前两个参数是强制性的,最后两个参数是可选的(具有未确定的参数数量)并且-d是可选标志.

Answer 1

argparse不喜欢metavar=''.

要么为它们命名,要么删除选项 - 但不要将它们留作空字符串.