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
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
Python x.x
res = True
while res:
res = input("Please confirm with y/yes...").lower(); res = res not in {'y','yes','Y','YES',''}
其他回答
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)
这个怎么样:
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
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()
这是我对它的看法,我只是想中止如果用户没有确认的行动。
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()
我修改了fmark的答案,用python 2/3兼容更pythonic。
如果您对更多错误处理感兴趣,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)