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
你可以尝试下面的代码来处理变量'accepted'中的选项:
print( 'accepted: {}'.format(accepted) )
# accepted: {'yes': ['', 'Yes', 'yes', 'YES', 'y', 'Y'], 'no': ['No', 'no', 'NO', 'n', 'N']}
这是密码。
#!/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
其他回答
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'
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
在2.7中,这是不是太非python化了?
if raw_input('your prompt').lower()[0]=='y':
your code here
else:
alternate code here
它至少能捕捉到“是”的任何变化。
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Alexander Artemenko提到的,这里有一个使用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')
使用
>>> 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