APT命令行界面 - 是/否输入？

Question

是否有任何简短的方法来实现APT(高级软件包工具)命令行界面在Python中的作用？

我的意思是,当包管理器提示后面是/否问题时[Yes/no],脚本接受YES/Y/yes/y或Enter(默认Yes为大写字母暗示).

我在官方文档找到的唯一的事情就是input和raw_input...

我知道这并不难以模仿,但重写是很烦人的:|

Answer 1

正如您所提到的,最简单的方法是使用raw_input()(或仅input()用于Python 3).没有内置的方法来做到这一点.来自食谱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 = raw_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")

用法示例:

>>> 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

Answer 2

我这样做:

# raw_input returns the empty string for "enter"
yes = {'yes','y', 'ye', ''}
no = {'no','n'}

choice = raw_input().lower()
if choice in yes:
   return True
elif choice in no:
   return False
else:
   sys.stdout.write("Please respond with 'yes' or 'no'")

Answer 3

strtoboolPython的标准库中有一个函数:http://docs.python.org/2/distutils/apiref.html？highlight = didutils.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.

Answer 4

一个非常简单(但不是非常复杂)的方法是单一选择:

msg = 'Shall I?'
shall = input("%s (y/N) " % msg).lower() == 'y'

您还可以编写一个简单的(略微改进的)函数:

def yn_choice(message, default='y'):
    choices = 'Y/n' if default.lower() in ('y', 'yes') else 'y/N'
    choice = input("%s (%s) " % (message, choices))
    values = ('y', 'yes', '') if choices == 'Y/n' else ('y', 'yes')
    return choice.strip().lower() in values

注意:在Python 2上,使用raw_input而不是input.

Answer 5

你可以使用click的confirm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Python 2/3于Linux,Mac或Windows.

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

Answer 6

正如@Alexander Artemenko所提到的,这是一个使用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')

#usage

>>> 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

Answer 7

我知道这已经回答了很多方法,这可能无法解决OP的具体问题(使用标准列表)但这是我为最常见的用例所做的,它比其他响应简单得多:

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

Answer 8

您也可以使用提示器.

无耻地从自述文件中获取:

#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

Answer 9

我修改了fmark的答案,通过python 2/3兼容更多pythonic.

如果您对具有更多错误处理的内容感兴趣,请参阅ipython的实用程序模块

# PY2/3 compatibility
from __future__ import print_function
# You could use the six package for this
try:
    input_ = raw_input
except NameError:
    input_ = input

def query_yes_no(question, default=True):
    """Ask a yes/no question via standard input and return the answer.

    If invalid input is given, the user will be asked until
    they acutally give valid input.

    Args:
        question(str):
            A question that is presented to the user.
        default(bool|None):
            The default value when enter is pressed with no value.
            When None, there is no default value and the query
            will loop.
    Returns:
        A bool indicating whether user has entered yes or no.

    Side Effects:
        Blocks program execution until valid input(y/n) is given.
    """
    yes_list = ["yes", "y"]
    no_list = ["no", "n"]

    default_dict = {  # default => prompt default string
        None: "[y/n]",
        True: "[Y/n]",
        False: "[y/N]",
    }

    default_str = default_dict[default]
    prompt_str = "%s %s " % (question, default_str)

    while True:
        choice = input_(prompt_str).lower()

        if not choice and default is not None:
            return default
        if choice in yes_list:
            return True
        if choice in no_list:
            return False

        notification_str = "Please respond with 'y' or 'n'"
        print(notification_str)

Answer 10

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")

该strtobool()函数将字符串转换为布尔值。如果无法解析字符串，则会引发 ValueError。

在 Python 3raw_input()中已重命名为input().

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

这是 的实现strtobool，如果你想把特殊词识别为true，你可以复制代码并添加你自己的案例。

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")

Answer 11

在 2.7 上，这是否太非 Pythonic 了？

if raw_input('your prompt').lower()[0]=='y':
   your code here
else:
   alternate code here

它至少捕获 Yes 的任何变化。