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

作为一个编程新手,我发现上面的一堆答案过于复杂,特别是如果目标是有一个简单的函数,你可以传递各种是/否问题,迫使用户选择是或否。在浏览了这篇文章和其他几篇文章,并借鉴了各种各样的好想法后,我得出了以下结论:

def yes_no(question_to_be_answered):
    while True:
        choice = input(question_to_be_answered).lower()
        if choice[:1] == 'y': 
            return True
        elif choice[:1] == 'n':
            return False
        else:
            print("Please respond with 'Yes' or 'No'\n")

#See it in Practice below 

musical_taste = yes_no('Do you like Pine Coladas?')
if musical_taste == True:
    print('and getting caught in the rain')
elif musical_taste == False:
    print('You clearly have no taste in music')

其他回答

对于Python 3,我使用这个函数:

def user_prompt(question: str) -> bool:
    """ Prompt the yes/no-*question* to the user. """
    from distutils.util import strtobool

    while True:
        user_input = input(question + " [y/n]: ")
        try:
            return bool(strtobool(user_input))
        except ValueError:
            print("Please use y/n or yes/no.\n")

函数的作用是:将字符串转换为bool类型。如果字符串不能被解析,它将引发ValueError。

在Python 3中,raw_input()已重命名为input()。

正如Geoff所说,strtoool实际上返回0或1,因此结果必须转换为bool类型。


这是strtobool的实现,如果你想让特殊的单词被识别为true,你可以复制代码并添加自己的case。

def strtobool (val):
    """Convert a string representation of truth to true (1) or false (0).
    True values are 'y', 'yes', 't', 'true', 'on', and '1'; false values
    are 'n', 'no', 'f', 'false', 'off', and '0'.  Raises ValueError if
    'val' is anything else.
    """
    val = val.lower()
    if val in ('y', 'yes', 't', 'true', 'on', '1'):
        return 1
    elif val in ('n', 'no', 'f', 'false', 'off', '0'):
        return 0
    else:
        raise ValueError("invalid truth value %r" % (val,))

你也可以使用提词器。

无耻地从README中截取:

#pip install prompter

from prompter import yesno

>>> yesno('Really?')
Really? [Y/n]
True

>>> yesno('Really?')
Really? [Y/n] no
False

>>> yesno('Really?', default='no')
Really? [y/N]
True

我知道这已经被回答了很多方法,这可能不能回答OP的具体问题(标准列表),但这是我为最常见的用例所做的,它比其他回答简单得多:

answer = input('Please indicate approval: [y/n]')
if not answer or answer[0].lower() != 'y':
    print('You did not indicate approval')
    exit(1)

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
def question(question, answers):
    acceptable = False
    while not acceptable:
        print(question + "specify '%s' or '%s'") % answers
        answer = raw_input()
        if answer.lower() == answers[0].lower() or answers[0].lower():
            print('Answer == %s') % answer
            acceptable = True
    return answer

raining = question("Is it raining today?", ("Y", "N"))

换做是我就会这么做。

输出

Is it raining today? Specify 'Y' or 'N'
> Y
answer = 'Y'