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
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
其他回答
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()
对python 3执行同样的操作。X, raw_input()不存在:
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")
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()
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