正如您所提到的，最简单的方法是使用raw_input()（或仅input()用于Python
3）。没有内置的方法可以做到这一点。来自配方
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代替input。） 用法示例：

>>> 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