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 3.8及以上版本的一行代码:
while res:= input("When correct, press enter to continue...").lower() not in {'y','yes','Y','YES',''}: pass
其他回答
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
这是我所使用的:
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
Python x.x
res = True
while res:
res = input("Please confirm with y/yes...").lower(); res = res not in {'y','yes','Y','YES',''}
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。
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()