Python虚拟咖啡店程序问题：输入无法匹配字典键，直接进入ELSE分支

核心问题 #

你的代码里，菜单字典的键是首字母大写格式（比如'Chicken Nuggets'），但你把用户输入转成了全小写（order.lower()），两者格式不匹配，导致order.lower() in menu永远为False，程序必然走else分支。

修复方案 #

需要在判断时将字典的键转为小写与输入匹配，同时保留原键格式，确保后续能正确从字典取值。修改while True循环内的代码如下：

while True:
    order = input("\nwhat would you like to order?\n")
    lower_order = order.lower()
    # 遍历菜单找到格式匹配的原键
    matched_item = None
    for item in menu:
        if item.lower() == lower_order:
            matched_item = item
            break
    if matched_item:
        otime = menu[matched_item]['cook_time']
        # 注意：otime是整数，拼接字符串前需转成str类型
        print_with_typing(f"\nThank you {name} I will be right out with your order, the {matched_item} will be ready in {str(otime)}\n\n")
        break
    else:
        time.sleep(1)
        print(f"\nSorry {name} we aren't serving {order}, please select one of the following items from our menu:\n")
        for item in menu:
            print(item)

额外修复点 #

原代码中输出烹饪时长时，otime是整数类型，直接和字符串拼接会触发类型错误，修复代码里已通过str(otime)解决该问题。

简化优化（可选） #

提前创建小写键到原键的映射字典，提升匹配效率：

# 定义menu后添加如下代码
lower_menu_map = {item.lower(): item for item in menu}

# 循环内判断逻辑简化为
while True:
    order = input("\nwhat would you like to order?\n")
    lower_order = order.lower()
    if lower_order in lower_menu_map:
        matched_item = lower_menu_map[lower_order]
        otime = menu[matched_item]['cook_time']
        print_with_typing(f"\nThank you {name} I will be right out with your order, the {matched_item} will be ready in {str(otime)}\n\n")
        break
    else:
        time.sleep(1)
        print(f"\nSorry {name} we aren't serving {order}, please select one of the following items from our menu:\n")
        for item in menu:
            print(item)

内容的提问来源于stack exchange，提问作者Apis