Question

我有一个简单的冰淇淋程序，一切正常。但是，如果输入无效，我该如何循环输入单个消息？当前，如果输入不正确，它将再次向用户询问第一个输入。意思是，如果我输入香草作为调味料，将勺数输入7，它将显示错误消息，然后，它不问您想要多少勺，而是询问您想要哪种调味料。在整个程序中，我该如何解决我的三个输入问题？输入有效订单后，我仍然希望程序重新启动。谢谢您的时间。

程序代码：

#Error for wrong flavor selection
class FlavorError(Exception):
    def __init__(self, flavors, message):
        Exception.__init__(self, flavors, message)
        self.flavors = flavors

#Error for wrong number of scoops selection
class ScoopsError(Exception):
    def __init__(self, scoops, message):
        Exception.__init__(self, scoops, message)
        self.scoops = scoops

#Error for wrong carrier selection
class HoldingsError(Exception):
    def __init__(self, holdings, message):
        Exception.__init__(self, holdings, message)
        self.holdings = holdings


def main():
    #Creating Values
    flavors= " "
    scoops = " "
    holdings = " "
    while True:
        try:
            #MENU / INFORMATION
            print("Flavors: vanilla, chocolate, strawberry, mint, pistachio, and spumoni")
            print("Number of scoops: 1, 2, 3")
            print("Ways to eat: bowl or cone")
            print()
            #FLAVOR INPUT
            flavors = input("Enter a flavor of ice cream: ").lower()
            print()
            #ERROR CHECK
            if flavors not in ['vanilla', 'chocolate', 'strawberry', 'mint', 'pistachio', 'spumoni']:
                raise FlavorError(flavors, "is not on the menu.")
            #NUMBER OF SCOOPS INPUT
            scoops = int(input("Enter the number of scoops: "))
            print()
            #ERROR CHECK
            if scoops > 3 or scoops <= 0:
                raise ScoopsError(scoops, 'We do not offer that many scoops!')
            #CARRIER INPUT
            holdings = input("Would you like a cone or bowl? ")
            print()
            #ERROR CHECK
            if holdings not in ['bowl', 'cone']:
                raise HoldingsError(holdings, 'Please select between a bowl or a cone')
            print(scoops , "scoops of" , flavors , "in a" , holdings)
            print()
        except Exception as e:
            print("Invalid choice! try again!"+str(e))
            print()

main()

Answer 1

有很多方法可以做到这一点，但是我相信这对您有用，并且您已经在使用类似的逻辑来实现自己的口味。

如果您需要更多的复杂性，则顶部的链接就是您想要的。

while scoops not in [1,2,3]:
    scoops = int(input("Enter the number of scoops: "))

Answer 2

要实现所需的行为，您可以在while循环中加入每个输入。

，仅在输入有效时才中断循环，这样它将继续执行程序

如何循环一个输入而不是重新启动整个程序？

2 个答案: