如何用循环实现汉堡点餐程序的重复运行？

要实现程序重复接受新用户点餐，用**while True无限循环**是最直接的方案，搭配退出判断就能灵活控制程序是否继续运行。

步骤说明： #

1. 将整个点餐流程（从欢迎语到打印感谢语）放入while True循环内部
2. 在打印完感谢语后，询问是否需要接待下一位顾客，根据输入决定继续循环或退出
3. 顺便修正原代码的逻辑错误：汉堡加芝士的判断误用了汉堡选择的变量，需改用芝士问题的输入结果

修改后的完整代码： #

def main():
    while True:
        print("Welcome to static burger!")
        print("Please answer each question with y or n.")

        # 账单与菜品价格初始化
        bill = 0.00
        hb = 8.00
        cb = 6.00
        cs = 5.00
        fry = 2.00
        cdmts = 1.00
        
        # 处理用户订单
        first = input("Would you like a hamburger? ")
        if first == 'y':
            cheese_input = input('with cheese? ')
            if cheese_input == 'y':
                bill += (hb + 2.00)
            else:
                bill += hb
        
        second = input("Would you like a cheeseburger? ")
        if second == 'y':
            bill += cb
        
        third = input("Would you like a chicken sandwich? ")
        if third == 'y':
            bill += cs
        
        fourth = input("Would you like fries? ")
        if fourth == 'y':
            bill += fry
        
        fifth = input("Would you like condiments? ")
        if fifth == 'y':
            bill += cdmts
        
        tip = 1.2
        total = bill * tip
        print("Your total is $ {:.2f} dollars".format(total))
        
        print("Thank you for your business!")
        
        # 询问是否继续接单
        continue_order = input("Would you like to take another order? (y/n) ")
        if continue_order.lower() != 'y':
            print("Closing the system...")
            break

if __name__ == "__main__":
    main()

关键细节： #

• while True会持续执行循环，直到遇到break语句终止
• 循环末尾的continue_order输入用来判断是否退出：若用户输入非'y'（忽略大小写），则执行break结束程序
• 每次循环开始时重新初始化bill为0，确保不同用户的订单金额互不干扰
• 修正了汉堡加芝士的逻辑判断，改用专门的输入变量cheese_input

内容的提问来源于stack exchange，提问作者sstatic