带有默认选项的python argparse选项

Question

我正在尝试在 Python 3 应用程序中使用 argparse，其中有一个明确的选择列表，但如果没有指定，则为默认值。

我的代码是:

parser.add_argument('--list', default='all', choices=['servers', 'storage', 'all'], help='list servers, storage, or both (default: %(default)s)') 
args = parser.parse_args()
print(vars(args))

但是，当我运行它时，我会得到以下选项:

$ python3 ./myapp.py --list all
{'list': 'all'}

或者没有选项:

$ python3 ./myapp.py --list
usage: myapp.py [-h] [--list {servers,storage,all}]
myapp.py: error: argument --list: expected one argument

我在这里遗漏了什么吗？或者我可以没有指定选项的默认值吗？

Answer 1

将 nargs 和 const 参数传递给 add_argument:

parser.add_argument('--list',
                    default='all',
                    const='all',
                    nargs='?',
                    choices=['servers', 'storage', 'all'],
                    help='list servers, storage, or both (default: %(default)s)')

如果您想知道 --list 是否在没有参数的情况下传递，请删除 const 参数，并检查 args.list是无。

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文档:

nargs用 '?'

One argument will be consumed from the command line if possible, and produced as a single item. If no command-line argument is present, the value from default will be produced. Note that for optional arguments, there is an additional case - the option string is present but not followed by a command-line argument. In this case the value from const will be produced.

const

When add_argument() is called with option strings (like -f or --foo) and nargs='?'. This creates an optional argument that can be followed by zero or one command-line arguments. When parsing the command line, if the option string is encountered with no command-line argument following it, the value of const will be assumed instead. See the nargs description for examples.