Question

我一直试图完成我的任务，但我遇到了逻辑错误。我使用的是Python 3。

print("Car Service Cost")
def main():
    loan=int(input("What is your loan cost?:\t"))
    maintenance=int(input("What is your maintenance cost?:\t"))
    total= loan + maintenance
    for rank in range(1,10000000):
         print("Total cost of Customer #",rank, "is:\t", total)
         checker()
def checker():
    choice=input("Do you want to proceed with the next customer?(Y/N):\t")
    if choice not in ["y","Y","n","N"]:
         print("Invalid Choice!")
    else:
         main()
main()

我得到了这个输出：

Car Service Cost
What is your loan cost?:    45
What is your maintenance cost?: 50
Total cost of Customer # 1 is:   95
Do you want to proceed with the next customer?(Y/N):    y
What is your loan cost?:    70
What is your maintenance cost?: 12
Total cost of Customer # 1 is:   82
Do you want to proceed with the next customer?(Y/N):    y
What is your loan cost?:    45
What is your maintenance cost?: 74
Total cost of Customer # 1 is:   119
Do you want to proceed with the next customer?(Y/N): here

我每次都排名为1。我做错了什么？

Answer 1

您不应该在main()中再次致电checker。您可以返回（如果您将其置于循环中，也可以使用break）：

def checker():
    while True:
        choice=input("Do you want to proceed with the next customer?(Y/N):\t")
        if choice not in ["y","Y","n","N"]:
            print("Invalid Choice!")
        else:
            return

如果您想在main输入'n'或'N'时突破循环，那么您可以尝试返回一个值：

def checker():
    while True:
        choice=input("Do you want to proceed with the next customer?(Y/N):\t")
        if choice not in ["y","Y","n","N"]:
            print("Invalid Choice!")
        else:
            return choice.lower()

然后检查'y'中是'n'还是main。

编辑：如果您不想使用return，您可以取出循环else，但这样您就无法检查用户是否想要停止：< / p>

def checker():
    choice=input("Do you want to proceed with the next customer?(Y/N):\t")
    if choice not in ["y","Y","n","N"]:
        print("Invalid Choice!")

Answer 2

您没有使用for循环。来自checker()，您再次致电main()。

而不是

else: main()

你应该只是return。

我不确定你在checker()做了你想做的事。

在函数中获取相同的范围值

2 个答案: