Python多函数游戏中如何复用用户名，避免重复询问并修复变量错误

错误原因 #

你在select_showroom()中写name = input，这是将内置函数input本身赋值给了name变量，而非用户在greet()中输入的用户名。所以f-string输出时会显示函数对象的默认描述<built_in_function_input>，这个问题和while循环无关。

解决方案：传递参数 #

核心思路是把greet()中获取的用户名作为参数，传递给后续需要使用该名称的函数，这样所有函数都能直接调用用户名，无需重复询问。

修改后的完整代码示例

def greet():
    # 修正原代码拼写错误：you → your
    name = input("what is your name?")
    
    print(f"hi there {name}") 
   
    answer = input("do you want to play a stupid game?")
    # 用lower()兼容大小写输入，elif让逻辑更严谨
    if answer.lower() == "yes":
        print("ok, let's play!")
        # 调用select_showroom时传入name参数
        select_showroom(name)
    elif answer.lower() == "no":
        print("well you're not missing anything!")
        exit()
    else:
        print("Please answer with 'yes' or 'no'.")
        # 错误输入后重新调用greet
        greet()

def select_showroom(name):
    # 接收name参数，直接在f-string中使用
    print(f"""{name}, you're in the lobby of the design building, three showrooms are open today.
          which one would you like to enter, 1, 2 or 3?""")
    choice = input("< ")
    
    if choice == "1":
        # 如果showroom1也需要用户名，同样传递参数
        showroom1(name)
    elif choice == "2":
        showroom2(name)
    elif choice == "3":
        showroom3(name)
    else:
        print("Invalid choice, try again.")
        # 错误输入后重新调用，保持传递name
        select_showroom(name)

# 示例后续函数，同样接收name参数
def showroom1(name):
    print(f"Welcome to Showroom 1, {name}!")

def showroom2(name):
    print(f"Welcome to Showroom 2, {name}!")

def showroom3(name):
    print(f"Welcome to Showroom 3, {name}!")

# 启动游戏
greet()

补充说明 #

• 所有需要使用用户名的后续函数（如showroom1、showroom2等），都需要在定义时添加name参数，并在调用时传入该参数。
• 原代码中的拼写错误（you→your）、逻辑漏洞（未处理yes/no以外的输入）也一并修正，提升游戏的健壮性。

内容的提问来源于stack exchange，提问作者Susan Young