Question

我正在尝试编写一个用户输入利率的简单代码块。该数字必须为0或更大，任何其他值将被拒绝，必须轮询用户，直到输入有效数字。如果该数字大于10％，则必须询问用户他/她是否真的期望利率高，如果用户回答是肯定的，则使用该号码，否则将要求用户输入价值再次，将进行上述检查。我无法理解嵌套循环方面的问题。非常感谢任何帮助！

def main():

    while True:
        try:
            interest_rate = int(input("Please enter an interest rate: "))
        except ValueErrror:
            print("Entered value is not a number! ")
        except KeyboardInterrupt:
            print("Command Error!")
        else:
            if 0 <= interest_rate < 10:
                break
            elif interest_rate > 10:
                print("Entered interest rate is greater than 10%. Are you sure? (y/n): ")

main()

Answer 1

在try if inp > 10中完成所有操作，询问用户是否满意并且如果他们是中断，elif用户输入在阈值内只是打破循环：

def main():
    while True:
        try:
            interest_rate = int(input("Please enter an interest rate: "))
            if interest_rate > 10:
                confirm = input("Entered interest rate is greater than 10%. Are you sure? (y/n): ")
                if confirm =="y":
                    break
            elif 0 <= interest_rate < 10:
                break
        except ValueError:
            print("Entered value is not a number! ") 
    return interest_rate

main()

Answer 2

有三件事情跳出来：

1）ValueErrror应为ValueError

2）您没有在最终测试中处理用户输入

3）您可能希望将< 10更改为<= 10

Answer 3

使print("Entered interest rate is greater than 10%. Are you sure? (y/n): ")成为输入

answer = int(input("Are you sure?"))
if answer == "y":
    break

Answer 4

else:
    if 0 <= interest_rate < 10:
        break
    elif interest_rate > 10:
        print("Entered interest rate is greater than 10%. Are you sure? (y/n): ")

可以是：

if 0 <= interest_rate <= 10:
    break
print("Entered interest rate is greater than 10%. Are you sure? (y/n): ")

除了最后一行必须接受响应并处理它。

您的else与if

无关

在elif

之后，您的break是不必要的

Answer 5

我通常更喜欢将解决方案和验证分解为不同的模块。请检查以下代码，看看我如何分解它们。因此，在调试和测试方面很容易。

def validating_user_input(num):
    """
    """
    return num > 0

def getting_user_input():
    """
    """
    user_input = int(raw_input("Enter the number that is greater than 0: "))
    return user_input

def confirming_choose():
    """
    """
    try:
        user_choose = int(raw_input("Can you confirm your input? [0|1]? "))
    except ValueError:
        return False
    return user_choose == 1



def main():
    """
    """
    initial_cond = True
    while initial_cond:
        user_input = getting_user_input()
        if validating_user_input(user_input):
            if user_input > 10:
                confirmation = confirming_choose()
                while not confirmation:
                    getting_user_input()
                #do you operating here
            initial_cond = False
        else:
            print "It is not valid input."


if __name__ == "__main__":
    main()

Python嵌套循环输入值并确认答案

5 个答案: