Argparse 需要一个参数 [英] Argparse expected one argument

问题描述

我有以下内容

import argparse

parser = argparse.ArgumentParser(prog='macc', usage='macc [options] [address]')
parser.add_argument('-l', '--list', help='Lists MAC Addresses')
args = parser.parse_args()
print(args)

def list_macs():
  print("Found the following MAC Addresses")

我在使用 python macc.py -l 运行时遇到错误，它说需要一个参数.即使我将代码更改为 parser.add_argument('-l', '--list', help='Lists MAC Addresses' default=1) 我也会遇到同样的错误.

I get an error when running with python macc.py -l it says that an argument was expected. Even when I change my code to parser.add_argument('-l', '--list', help='Lists MAC Addresses' default=1) I get the same error.

推荐答案

参数的默认操作是 store，它设置 parser 返回的命名空间中的属性值.parse_args 使用下一个命令行参数.

The default action for an argument is store, which sets the value of the attribute in the namespace returned by parser.parse_args using the next command line argument.

您不想存储任何特定值；您只想确认使用了 -l.一个快速的技巧是使用 store_true 操作(这会将 args.list 设置为 True).

You don't want to store any particular value; you just want to acknowledge that -l was used. A quick hack would be to use the store_true action (which would set args.list to True).

parser = argparse.ArgumentParser(prog='macc')
parser.add_argument('-l', '--list', action='store_true', help='Lists MAC Addresses')

args = parser.parse_args()

if args.list:
    list_macs()

store_true 操作也意味着 type=bool 和 default=False.

The store_true action implies type=bool and default=False as well.

然而，一个稍微简洁的方法是定义一个名为 list 的子命令.使用这种方法，您的调用将是 macc.py list 而不是 macc.py --list.

However, a slightly cleaner approach would be to define a subcommand named list. With this approach, your invocation would be macc.py list rather than macc.py --list.

parser = argparse.ArgumentParser(prog='macc')
subparsers = parser.add_subparsers(dest='cmd_name')
subparsers.add_parser('list')

args = parser.parse_args()

if args.cmd_name == "list":
    list_macs()

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