有没有什么简单的方法来实现APT(高级包工具)命令行界面在Python中的作用?

我的意思是,当包管理器提示一个yes/no问题,后面跟着[yes/no]时,脚本接受yes/ Y/yes/ Y或Enter(默认为yes,由大写字母提示)。

我在官方文档中唯一找到的是input和raw_input…

我知道模仿它并不难,但是重写:|很烦人


当前回答

正如Alexander Artemenko提到的,这里有一个使用strtobool()的简单解决方案。

from distutils.util import strtobool

def user_yes_no_query(question):
    sys.stdout.write('%s [y/n]\n' % question)
    while True:
        try:
            return strtobool(raw_input().lower())
        except ValueError:
            sys.stdout.write('Please respond with \'y\' or \'n\'.\n')

使用

>>> user_yes_no_query('Do you like cheese?')
Do you like cheese? [y/n]
Only on tuesdays
Please respond with 'y' or 'n'.
ok
Please respond with 'y' or 'n'.
y
>>> True

其他回答

由于答案是“是”或“否”,在下面的例子中,第一个解决方案是使用while函数重复这个问题,第二个解决方案是使用递归-是定义事物本身的过程。

def yes_or_no(question):
    while "the answer is invalid":
        reply = str(input(question+' (y/n): ')).lower().strip()
        if reply[:1] == 'y':
            return True
        if reply[:1] == 'n':
            return False

yes_or_no("Do you know who Novak Djokovic is?")

第二个解决方案:

def yes_or_no(question):
    """Simple Yes/No Function."""
    prompt = f'{question} ? (y/n): '
    answer = input(prompt).strip().lower()
    if answer not in ['y', 'n']:
        print(f'{answer} is invalid, please try again...')
        return yes_or_no(question)
    if answer == 'y':
        return True
    return False

def main():
    """Run main function."""
    answer = yes_or_no("Do you know who Novak Djokovic is?")
    print(f'you answer was: {answer}')


if __name__ == '__main__':
    main()

在Python的标准库中有一个函数strtoool: http://docs.python.org/2/distutils/apiref.html?highlight=distutils.util#distutils.util.strtobool

您可以使用它来检查用户的输入并将其转换为True或False值。

您可以使用单击的确认方法。

import click

if click.confirm('Do you want to continue?', default=True):
    print('Do something')

这将打印:

$ Do you want to continue? [Y/n]:

应该适用于Linux, Mac或Windows上的Python 2/3。

文档:http://click.pocoo.org/5/prompts/ # confirmation-prompts

正如你提到的,最简单的方法是使用raw_input()(或简单的input()对于Python 3)。没有内置的方法可以做到这一点。配方577058:

import sys


def query_yes_no(question, default="yes"):
    """Ask a yes/no question via raw_input() and return their answer.

    "question" is a string that is presented to the user.
    "default" is the presumed answer if the user just hits <Enter>.
            It must be "yes" (the default), "no" or None (meaning
            an answer is required of the user).

    The "answer" return value is True for "yes" or False for "no".
    """
    valid = {"yes": True, "y": True, "ye": True, "no": False, "n": False}
    if default is None:
        prompt = " [y/n] "
    elif default == "yes":
        prompt = " [Y/n] "
    elif default == "no":
        prompt = " [y/N] "
    else:
        raise ValueError("invalid default answer: '%s'" % default)

    while True:
        sys.stdout.write(question + prompt)
        choice = input().lower()
        if default is not None and choice == "":
            return valid[default]
        elif choice in valid:
            return valid[choice]
        else:
            sys.stdout.write("Please respond with 'yes' or 'no' " "(or 'y' or 'n').\n")

(对于Python 2,使用raw_input而不是input。) 使用的例子:

>>> query_yes_no("Is cabbage yummier than cauliflower?")
Is cabbage yummier than cauliflower? [Y/n] oops
Please respond with 'yes' or 'no' (or 'y' or 'n').
Is cabbage yummier than cauliflower? [Y/n] [ENTER]
>>> True

>>> query_yes_no("Is cabbage yummier than cauliflower?", None)
Is cabbage yummier than cauliflower? [y/n] [ENTER]
Please respond with 'yes' or 'no' (or 'y' or 'n').
Is cabbage yummier than cauliflower? [y/n] y
>>> True

正如Alexander Artemenko提到的,这里有一个使用strtobool()的简单解决方案。

from distutils.util import strtobool

def user_yes_no_query(question):
    sys.stdout.write('%s [y/n]\n' % question)
    while True:
        try:
            return strtobool(raw_input().lower())
        except ValueError:
            sys.stdout.write('Please respond with \'y\' or \'n\'.\n')

使用

>>> user_yes_no_query('Do you like cheese?')
Do you like cheese? [y/n]
Only on tuesdays
Please respond with 'y' or 'n'.
ok
Please respond with 'y' or 'n'.
y
>>> True