Question

我需要一个能够排除错误输入并再次要求输入的功能。但在输入错误后，它会返回None而不是新输入。我的代码有什么问题，我该如何解决？

def start():

    def inputNumber(answer):
        try:
            number = int(input(answer))
            if number <= 100 and number >= 0:
                print('%%%',number,'%%%')
                return number
            else:
                inputNumber('Number is wrong, please input number from 0 to 100: ')
        except (ValueError):
            inputNumber('It is not a number, please input number from 0 to 100: ')

    def checkInput(number2):
        print('$$$',number2,'$$$')
        if number < 50:
            return number2
        else:
            return checkInput(inputNumber('Input number from 0 to 100: '))

    number = 0
    print('###',checkInput(inputNumber('Input number from 0 to 100: ')),'###')
    start()

start()

结果如下：

输入数字0到100：777

数字错误，请输入0到100之间的数字：sadf

不是数字，请输入0到100：17之间的数字

%%% 17 %%%

$$$无$$$

TypeError：unorderable类型：NoneType（）＆lt; INT（）

Answer 1

您正在递归地调用inputNumber但不会返回递归调用的结果。更好地使用循环而不是递归：

def inputNumber(prompt):
    while True:
        try:
            number = int(input(prompt))
            if 0 <= number <= 100:
                print('%%%',number,'%%%')
                return number
            prompt = 'Number is wrong, please input number from 0 to 100: '
        except ValueError:
            prompt = 'It is not a number, please input number from 0 to 100: '

顺便说一句：您还应该在其他功能中使用循环，而不是定义嵌套函数。

Answer 2

问题是，一旦检查失败，您再次致电inputNumber，但不对答案做任何事情。你需要退货。

def start():

    def inputNumber(answer):
        try:
            number = int(input(answer))
            if number <= 100 and number >= 0:
                print('%%%', number, '%%%')
                return number
            else:
                return inputNumber('Number is wrong, please input number from 0 to 100: ')
        except (ValueError):
            return inputNumber('It is not a number, please input number from 0 to 100: ')

    def checkInput(number2):
        print('$$$', number2, '$$$')
        if number < 50:
            return number2
        else:
            return checkInput(inputNumber('Input number from 0 to 100: '))

    number = 0
    print('###', checkInput(inputNumber('Input number from 0 to 100: ')), '###')
    start()

start()

Answer 3

运行此代码它将解决您的问题。你只需要捕获额外的错误（NameError）

def start():

    def inputNumber(answer):
        try:
            number = int(input(answer))
            if number <= 100 and number >= 0:
                print('%%%',number,'%%%')
                return number
            else:
                inputNumber('Number is wrong, please input number from 0 to 100: ')
        except (ValueError, NameError) as e:
            inputNumber('It is not a number, please input number from 0 to 100: ')

    def checkInput(number2):
        print('$$$',number2,'$$$')
        if number < 50:
            return number2
        else:
            return checkInput(inputNumber('Input number from 0 to 100: '))

    number = 0
    print('###',checkInput(inputNumber('Input number from 0 to 100: ')),'###')
    start()

start()

错误输入后，函数返回None

3 个答案: