예 / 아니요 입력과 같은 APT 명령 행 인터페이스?

예 / 아니요 입력과 같은 APT 명령 행 인터페이스?

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Python에서 APT ( Advanced Package Tool ) 명령 행 인터페이스의 기능을 수행 할 수있는 짧은 방법이 있습니까?

패키지 관리자가 예 / 아니오 질문 다음에를 묻는 메시지를 표시 [Yes/no]하면 스크립트가 YES/Y/yes/y또는 Enter(기본값 Yes은 대문자로 표시됨) 을 수락 한다는 의미 입니다.

나는 공식 문서에서 찾을 수있는 유일한 방법입니다 input및 raw_input...

나는 모방하기가 어렵지 않다는 것을 알고 있지만 다시 작성하는 것은 성가신 일입니다. |

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당신이 언급 한 바와 같이, 가장 쉬운 방법은 사용하는 것입니다 raw_input()(또는 단순히 input()위한 파이썬 3 ). 이 작업을 수행하는 기본 방법은 없습니다. 에서 레시피 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")

사용 예 :

>>> 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

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나는 이렇게 할 것입니다 :

# 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'")

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strtobool파이썬의 표준 라이브러리 에는 다음과 같은 함수 가 있습니다 : http://docs.python.org/2/distutils/apiref.html?highlight=distutils.util#distutils.util.strtobool

이를 사용하여 사용자 입력을 확인하고 값 True또는 False값으로 변환 할 수 있습니다.

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단일 선택을 위해이 작업을 수행하는 매우 간단하지만 (정교하지는 않은) 방법은 다음과 같습니다.

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

이 주위에 간단한 (약간 개선 된) 함수를 작성할 수도 있습니다.

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

참고 : Python 2에서는 raw_input대신을 사용하십시오 input.

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click 의 confirm방법을 사용할 수 있습니다 .

import click

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

인쇄됩니다 :

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

Python 2/3Linux, Mac 또는 Windows에서 작동합니다 .

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

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@Alexander Artemenko가 언급했듯이 strtobool을 사용하는 간단한 솔루션이 있습니다.

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

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나는 이것이 여러 가지 방법으로 답변되었으며 OP의 특정 질문 (기준 목록과 함께)에 대답하지 않을 수도 있지만 이것이 가장 일반적인 유스 케이스에 대해 한 일이며 다른 답변보다 훨씬 간단합니다.

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

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prompter를 사용할 수도 있습니다 .

README에서 뻔뻔스럽게 가져온 것 :

#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

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python 2/3 호환 더 많은 pythonic으로 fmark의 답변을 수정했습니다.

더 많은 오류 처리에 관심이 있다면 ipython의 유틸리티 모듈을 참조하십시오.

# 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)

---

2.7에서, 이것은 비파이 토닉입니까?

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

적어도 예의 변형을 포착합니다.

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raw_input()존재하지 않는 python 3.x와 동일하게 수행하십시오 .

def ask(question, default = None):
    hasDefault = default is not None
    prompt = (question 
               + " [" + ["y", "Y"][hasDefault and default] + "/" 
               + ["n", "N"][hasDefault and not default] + "] ")

    while True:
        sys.stdout.write(prompt)
        choice = input().strip().lower()
        if choice == '':
            if default is not None:
                return default
        else:
            if "yes".startswith(choice):
                return True
            if "no".startswith(choice):
                return False

        sys.stdout.write("Please respond with 'yes' or 'no' "
                             "(or 'y' or 'n').\n")

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아래 코드와 같은 것을 시도하여 변수 'accepted'표시에서 선택 항목을 사용할 수 있습니다.

print( 'accepted: {}'.format(accepted) )
# accepted: {'yes': ['', 'Yes', 'yes', 'YES', 'y', 'Y'], 'no': ['No', 'no', 'NO', 'n', 'N']}

코드는 다음과 같습니다 ..

#!/usr/bin/python3

def makeChoi(yeh, neh):
    accept = {}
    # for w in words:
    accept['yes'] = [ '', yeh, yeh.lower(), yeh.upper(), yeh.lower()[0], yeh.upper()[0] ]
    accept['no'] = [ neh, neh.lower(), neh.upper(), neh.lower()[0], neh.upper()[0] ]
    return accept

accepted = makeChoi('Yes', 'No')

def doYeh():
    print('Yeh! Let\'s do it.')

def doNeh():
    print('Neh! Let\'s not do it.')

choi = None
while not choi:
    choi = input( 'Please choose: Y/n? ' )
    if choi in accepted['yes']:
        choi = True
        doYeh()
    elif choi in accepted['no']:
        choi = True
        doNeh()
    else:
        print('Your choice was "{}". Please use an accepted input value ..'.format(choi))
        print( accepted )
        choi = None

---

As a programming noob, I found a bunch of the above answers overly complex, especially if the goal is to have a simple function that you can pass various yes/no questions to, forcing the user to select yes or no. After scouring this page and several others, and borrowing all of the various good ideas, I ended up with the following:

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')

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For Python 3, I'm using this function:

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]: ").lower()
        try:
            result = strtobool(user_input)
            return result
        except ValueError:
            print("Please use y/n or yes/no.\n")

The strtobool function converts a string into a bool. If the string cant be parsed it will raise a ValueError.

In Python 3 raw_input has been renamed to input.

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How about this:

def yes(prompt = 'Please enter Yes/No: '):
while True:
    try:
        i = raw_input(prompt)
    except KeyboardInterrupt:
        return False
    if i.lower() in ('yes','y'): return True
    elif i.lower() in ('no','n'): return False

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This is what I use:

import sys

# cs = case sensitive
# ys = whatever you want to be "yes" - string or tuple of strings

#  prompt('promptString') == 1:               # only y
#  prompt('promptString',cs = 0) == 1:        # y or Y
#  prompt('promptString','Yes') == 1:         # only Yes
#  prompt('promptString',('y','yes')) == 1:   # only y or yes
#  prompt('promptString',('Y','Yes')) == 1:   # only Y or Yes
#  prompt('promptString',('y','yes'),0) == 1: # Yes, YES, yes, y, Y etc.

def prompt(ps,ys='y',cs=1):
    sys.stdout.write(ps)
    ii = raw_input()
    if cs == 0:
        ii = ii.lower()
    if type(ys) == tuple:
        for accept in ys:
            if cs == 0:
                accept = accept.lower()
            if ii == accept:
                return True
    else:
        if ii == ys:
            return True
    return False

---

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"))

This is how I'd do it.

Output

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

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Here's my take on it, I simply wanted to abort if the user did not affirm the action.

import distutils

if unsafe_case:
    print('Proceed with potentially unsafe thing? [y/n]')
    while True:
        try:
            verify = distutils.util.strtobool(raw_input())
            if not verify:
                raise SystemExit  # Abort on user reject
            break
        except ValueError as err:
            print('Please enter \'yes\' or \'no\'')
            # Try again
    print('Continuing ...')
do_unsafe_thing()

참고URL : https://stackoverflow.com/questions/3041986/apt-command-line-interface-like-yes-no-input