为什么即使输入是整数，我也会不断收到 TypeError？

Question

我正在为一个学校项目编写一些代码，该项目要求用户输入一个介于 0-10 之间的数字，我不知道为什么即使使用列表将输入转换为 integer，我仍然会收到 TypeError。 请问有什么帮助吗？ 我在 Wing IDE 101 上执行此操作。

numbers = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
while True:
    num = ''
    while num == '':
        try:
            num = input("Please enter a number: ")
            num = int(num)
        except ValueError:
            if num in numbers:
                num = numbers.index(num)
                num = int(num)
            else:
                print('Invalid input, please enter valid input')
    if num == 0:
        print("Thank you for using this program")
        break
    elif num < 0 or num > 10:
        print("This is an invalid number, please enter a valid number.")
        print()
        continue

Answer 1

我觉得是这个原因：

当您的输入不是像"asdf"这样的有效输入时。 它会跳出内部的while循环。 因为你内部的 while 循环条件是num == ''而num现在是"asdf" 。

您只需将num = ""放在print('Invalid input, please enter valid input')下，您的代码就可以正常工作并且不再引发 TypeError。

完整代码如下：

numbers = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
while True:
    num = ''
    while num == '':
        try:
            num = input("Please enter a number: ")
            num = int(num)
        except ValueError:
            if num in numbers:
                num = numbers.index(num)
                num = int(num)
            else:
                print('Invalid input, please enter valid input')
                num = ""
    if num == 0:
        print("Thank you for using this program")
        break
    elif num < 0 or num > 10:
        print("This is an invalid number, please enter a valid number.")
        print()
        continue

实际上，您的代码可以简化为一个循环：

numbers = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
while True:
    num = input("Please enter a number: ")
    if num in numbers:
        num = numbers.index(num)
    else:
        try:
            num = int(num)
        except ValueError:
            print("Invalid input, please enter valid input")
            continue
    if num == 0:
        print("Thank you for using this program")
        break
    elif num < 0 or num > 10:
        print("This is an invalid number, please enter a valid number.")

Answer 2

发生在你身上的事情在动态语言中可能会发生很多，因为类型是在运行时而不是在设计时解析的，所以有时很难确定。

首先，将user_input与您正在评估的数字 ( num ) 分离为 2 个不同的参数（一个用于用户输入字符串，一个用于整数） - 我知道在 Python 上并不重要，但这是一个很好的做法。

现在，您的num值只会在成功解析时分配。 在循环结束时也不需要continue

代码片段

numbers = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"]
while True:
    user_input = ''
    num = None  # default invalid value
    while num is None:
        try:
            user_input = input("Please enter a number: ")
            num = int(user_input)
        except ValueError:
            if user_input in numbers:
                num = numbers.index(user_input)
            else:
                print('Invalid input, please enter valid input')
                print()
    if num == 0:
        print("Thank you for using this program")
        break
    elif num < 0 or num > 10:
        print("This is an invalid number, please enter a valid number.")
        print()

调试Output

Please enter a number: 1
Please enter a number: one
Please enter a number: 11
This is an invalid number, please enter a valid number.

Please enter a number: -1
This is an invalid number, please enter a valid number.

Please enter a number: no numbder
Invalid input, please enter valid input

Please enter a number: 0
Thank you for using this program

Process finished with exit code 0