Codecademy循环咖啡聊天机器人：退出词功能异常求助

我是Codecademy平台《Build Chatbots with Python》技能路径的初学者，正在完成Looping Coffee Chatbot项目的第13步附加任务——给聊天机器人添加识别"stop""bye"等退出词的功能，要求在任意步骤输入这些词都能终止订单。但现在输入退出词时，不会触发退出逻辑（打印"Sure, feel free to come back anytime!"并终止函数），反而会调用utils.py里的print_message函数。

相关代码如下：

script.py #

from utils import print_message, get_size, order_latte
exit_words = ["stop", "bye"]

def coffee_bot():
  print('Welcome to the cafe!')
  order_drink = "y"
  drinks = []
  while order_drink in  ["y", "yes", "affirmative", "amen", "fine", "good", "okay", "true", "yea"]:
    size = get_size()
    if size in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    drink_type = get_drink_type()
    if drink_type in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    drink = '{} {}'.format(size, drink_type)
    print('Alright, that\'s a {}!'.format(drink))
    drinks.append(drink)
    while True:
      order_drink = input("Would you like to order another drink? (y/n) \n> ")
      if order_drink in ["y", "n", "yes", "affirmative", "amen", "fine", "good", "okay", "true", "yea", "nah", "nay", "nix"]:
        break
      elif order_drink in exit_words:
        print("Sure, feel free to come back anytime!")
        return
    print("Okay, so I have:")
    for drink in drinks:
      print("-", drink)
    name = input('Can I get your name please? \n> ')
    if name in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    print('Thanks, {}! Your order will be ready shortly.'.format(name))

def get_drink_type():
  res = input('What type of drink would you like? \n[a] Brewed Coffee \n[b] Mocha \n[c] Latte \n> ')

  if res == 'a':
    return 'brewed coffee'
  elif res == 'b':
    return order_mocha()
  elif res == 'c':
    return order_latte()
  else:
    print_message()
    return get_drink_type()
  
# Define new functions here!
def order_mocha():
  while True:
    res = input("Would you like to try our limited-edition peppermint mocha? \n[a] Sure! \n[b] Maybe next time! \n> ")
    if res == "a":
      return "peppermint mocha"
    elif res == "b":
      return "mocha"
    print_message()

coffee_bot()

utils.py #

exit_words = ["stop", "bye"]
def print_message():
  print('I\'m sorry, I did not understand your selection. Please enter the corresponding letter for your response.')
  
def get_size():
  res = input('What size drink can I get for you? \n[a] Small \n[b] Medium \n[c] Large \n> ')
  
  if res == 'a':
    return 'small'
  elif res == 'b':
    return 'medium'
  elif res == 'c':
    return 'large'
  else:
    print_message()
    return get_size()

def order_latte():
  res = input('And what kind of milk for your latte? \n[a] 2% milk \n[b] Non-fat milk \n[c] Soy milk \n> ')

  if res == 'a':
    return 'latte'
  elif res == 'b':
    return 'non-fat latte'
  elif res == 'c':
    return 'soy latte'
  else:
    print_message()
    return order_latte()

所有处理输入的函数（get_size、get_drink_type、order_mocha、order_latte）都没有优先检查输入是否为退出词，而是直接将非指定字母的输入判定为无效，调用print_message并递归重试。这导致退出词根本无法返回给上层的coffee_bot函数，自然触发不了退出逻辑。

在每个处理输入的函数中，先判断输入是否属于退出词，如果是则直接返回该词，让上层的coffee_bot能捕获并执行退出逻辑。同时统一用lower()处理输入，兼容大小写差异。

修改后的代码如下：

修改后的utils.py #

exit_words = ["stop", "bye"]
def print_message():
  print('I\'m sorry, I did not understand your selection. Please enter the corresponding letter for your response.')
  
def get_size():
  res = input('What size drink can I get for you? \n[a] Small \n[b] Medium \n[c] Large \n> ')
  # 优先检查退出词
  if res.lower() in exit_words:
    return res.lower()
  if res == 'a':
    return 'small'
  elif res == 'b':
    return 'medium'
  elif res == 'c':
    return 'large'
  else:
    print_message()
    return get_size()

def order_latte():
  res = input('And what kind of milk for your latte? \n[a] 2% milk \n[b] Non-fat milk \n[c] Soy milk \n> ')
  # 优先检查退出词
  if res.lower() in exit_words:
    return res.lower()
  if res == 'a':
    return 'latte'
  elif res == 'b':
    return 'non-fat latte'
  elif res == 'c':
    return 'soy latte'
  else:
    print_message()
    return order_latte()

修改后的script.py #

from utils import print_message, get_size, order_latte
exit_words = ["stop", "bye"]

def coffee_bot():
  print('Welcome to the cafe!')
  order_drink = "y"
  drinks = []
  while order_drink in  ["y", "yes", "affirmative", "amen", "fine", "good", "okay", "true", "yea"]:
    size = get_size()
    if size in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    drink_type = get_drink_type()
    if drink_type in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    drink = '{} {}'.format(size, drink_type)
    print('Alright, that\'s a {}!'.format(drink))
    drinks.append(drink)
    while True:
      order_drink = input("Would you like to order another drink? (y/n) \n> ")
      # 优先检查退出词
      if order_drink.lower() in exit_words:
        print("Sure, feel free to come back anytime!")
        return
      if order_drink.lower() in ["y", "n", "yes", "affirmative", "amen", "fine", "good", "okay", "true", "yea", "nah", "nay", "nix"]:
        order_drink = order_drink.lower()
        break
      else:
        print_message()
    # 用户选择不再点单时终止循环
    if order_drink == 'n':
      break
    print("Okay, so I have:")
    for drink in drinks:
      print("-", drink)
    name = input('Can I get your name please? \n> ')
    if name.lower() in exit_words:
      print("Sure, feel free to come back anytime!")
      return
    print('Thanks, {}! Your order will be ready shortly.'.format(name))

def get_drink_type():
  res = input('What type of drink would you like? \n[a] Brewed Coffee \n[b] Mocha \n[c] Latte \n> ')
  # 优先检查退出词
  if res.lower() in exit_words:
    return res.lower()
  if res == 'a':
    return 'brewed coffee'
  elif res == 'b':
    return order_mocha()
  elif res == 'c':
    return order_latte()
  else:
    print_message()
    return get_drink_type()
  
def order_mocha():
  while True:
    res = input("Would you like to try our limited-edition peppermint mocha? \n[a] Sure! \n[b] Maybe next time! \n> ")
    # 优先检查退出词
    if res.lower() in exit_words:
      return res.lower()
    if res == "a":
      return "peppermint mocha"
    elif res == "b":
      return "mocha"
    print_message()

coffee_bot()

内容的提问来源于stack exchange，提问作者Noah