python 3 argparse调用函数 [英] python 3 argparse call a function

问题描述

我想在python3中创建类似命令行/类似shell的界面.

I wanted to create a commandline-like / shell-like interface in python3.

Argparse似乎可以解析和显示帮助/错误消息.根据 argparse的python3文档，有一个 func = 参数，该参数可用于获取由argparse调用的函数.

Argparse seems to do the job of parsing and displaying the help/error messages. According to the python3 documentation of argparse, there is a func= argument that can be used to get your function called by argparse.

# sub-command functions
def foo(args):
   print(args.x * args.y)
def bar(args):
    print('((%s))' % args.z)
# create the top-level parser
parser = argparse.ArgumentParser()
subparsers = parser.add_subparsers()
# create the parser for the "foo" command
parser_foo = subparsers.add_parser('foo')
parser_foo.add_argument('-x', type=int, default=1)
parser_foo.add_argument('y', type=float)
parser_foo.set_defaults(func=foo)
# create the parser for the "bar" command
parser_bar = subparsers.add_parser('bar')
parser_bar.add_argument('z')
parser_bar.set_defaults(func=bar)

但是据我所知help_parser.set_defaults(func=foo)并不称呼我的功能.如果您能帮助我，将不胜感激.

But as far as I can tell help_parser.set_defaults(func=foo) does not call my funciton. It would be appreciated if you could help me.

您可以通过使用python3运行程序，键入help，然后按[Enter]来重现该问题.它不会按预期方式打印hello. 谢谢！

You can reproduce the issue by running the program with python3, typing help and then press [Enter]. It does not print hello as expected. Thanks!

def foo():
    print('hello')


class Console:
    def __init__(self):
        """Console like interface for navigating and interacting with the external file system."""
        parser = argparse.ArgumentParser(
            description='Console like interface for navigating and interacting with the external file system.',
            add_help=False)
        subparsers = parser.add_subparsers(dest='command')
        subparsers.required = True

        help_parser = subparsers.add_parser('help')
        help_parser.add_argument('-x', type=int, default=1)
        help_parser.set_defaults(func=foo)

        setting_parser = subparsers.add_parser('settings')
        setting_subparsers = setting_parser.add_subparsers(dest='settings_command')
        setting_subparsers.required = True

        setting_save_parser = setting_subparsers.add_parser('save', help='Save the current settings in a .json file.')
        setting_save_parser.add_argument('file', type=str, nargs='?', default='settings.json')

        setting_load_parser = setting_subparsers.add_parser('load', description='Load the last settings from a .json file.')
        setting_load_parser.add_argument('file', type=str, nargs='?', help='settings.json')

        setting_set_parser = setting_subparsers.add_parser('set')
        setting_set_parser.add_argument('--host', type=str, required=True)
        setting_set_parser.add_argument('-u', '--username', type=str, required=True)
        setting_set_parser.add_argument('-p', '--password', type=str, required=True)
        setting_set_parser.add_argument('-x', '--proxy', type=str, required=False)

        while True:
            try:
                print('', flush=True, sep='')
                data = input('>>>').split(' ')
                print('your command:', data)
                parser.parse_args(data)
            except SystemExit:
                pass

if __name__ == '__main__':
    """Spawn an commandline like interface."""
    c = Console()

推荐答案

实际上argparse不会自动调用该方法-您必须自己执行.它所做的全部将方法添加到args的func属性中.因此，您可以做的是检查func属性是否存在，然后按以下方式调用它:

Actually argparse won't call the method automatically - you have to do it yourself. All it does it adds the method to the func attribute of the args. So, what you could do is to check if func attribute exists and then invoke it as following:

args = parser.parse_args(data)
if hasattr(args, 'func'):
    args.func()

这篇关于python 3 argparse调用函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！