猜数字游戏的输入验证导致错误

Question

这是我在这个社区的第一篇文章，当然我是初学者。我期待着有一天我可以帮助别人。无论如何，这是一个简单的代码，我想要它，以便在用户输入字符串时出现错误。不幸的是，它没有按照我想要的方式执行，这是代码：

number = 1

guess = int(input('Guess this number: '))

while True:
    try:
        if guess > number:
            print("Number is too high, go lower, try again")
            guess = int(input('Guess this number: '))
        elif guess < number:
            print("Too low, go higher, try again")
            guess = int(input('Guess this number: '))
        else:
            print("That is correct")
            break
    except (SyntaxError, ValueError):
            print("You can only enetr numbers, try again")

当程序执行时，它要求我“猜猜这个数字：”，当我写任何字符串时，例如“d”，它给出了错误：

Guess this number: d
Traceback (most recent call last):
  File "Numberguess.py", line 5, in <module>
    guess = int(input('Guess this number: '))
ValueError: invalid literal for int() with base 10: 'd'

---

感谢您的时间和支持。

Answer 1

问题出在第一行。当您将用户的输入直接转换为int时，当用户输入字母时，该字母无法转换为int，从而导致您收到错误消息。 要解决这个问题，你需要做

guess = input('Guess this number: ')
if not guess.isdigit():
    raise ValueError("input must be of type int")

Answer 2

我只会使用.isdigit()。该字符串将在该点进行验证，然后如果验证有效，您会将其转换为 int。

guess = input('Guess this number: ')
if guess.isdigit():
    guess = int(guess)
else:
    print("You can only enter numbers, try again")

还值得一提的是，try/excepts 很酷，他们可以完成工作，但尝试将它们减少到零是一个好习惯，而不是捕获错误，而是事先验证数据。

下一个例子会这样做：

number = 1

while True:

    # Step 1, validate the user choice
    guess = input('Guess this number: ')
    if guess.isdigit():
        guess = int(guess)
    else:
        print("You can only enter numbers, try again")
        continue

    # Step 2, play the guess game
    if guess > number:
        print("Number is too high, go lower, try again")
    elif guess < number:
        print("Too low, go higher, try again")
    else:
        print("That is correct")
        break

Answer 3

看看这一行：

guess = int(input('Guess this number: '))

您尝试将字符串转换为 int，这是可能的，但前提是字符串表示数字。

这就是你得到错误的原因。 except 对您不起作用，因为您从“try”中获得了变量的输入。

顺便说一句，没有理由在if中输入，所以你的代码应该是这样的：

number = 1

while True:
    try:
        guess = int(input('Guess this number: '))

        if guess > number:
            print("Number is too high, go lower, try again")
        elif guess < number:
            print("Too low, go higher, try again")
        else:
            print("That is correct")
            break
    except (SyntaxError, ValueError):
            print("You can only enetr numbers, try again")

Answer 4

欢迎来到堆栈溢出！每个人都需要从某个地方开始

看看下面的代码：

# import random to generate a random number within a range
from random import randrange


def main():
    low = 1
    high = 100

    # gen rand number
    number = gen_number(low, high)

    # get initial user input
    guess = get_input()

    # if user did not guess correct than keep asking them to guess
    while guess != number:
        if guess > number:
            print("You guessed too high!")
            guess = get_input()
        if guess < number:
            print("You guess too low!")
            guess = get_input()
    # let the user know they guess correct
    print(f"You guessed {guess} and are correct!")


def gen_number(low, high):
    return randrange(low, high)


# function to get input from user
def get_input():

    guess = input(f"Guess a number (q to quit): ")
    if guess == 'q':
        exit()

    # check to make sure user input is a number otherwise ask the user to guess again
    try:
        guess = int(guess)
    except ValueError:
        print("Not a valid number")
        get_input()

    # return guess if it is a valid number
    return guess


# Main program
if __name__ == '__main__':
    main()

这是一个包含 Python 随机模块以在一个范围内生成随机数的绝佳机会。还有一个很好的递归示例，不断要求用户猜测，直到他们提供有效输入。如果可行，请将此答案标记为正确，如果您有任何问题，请随时离开 cmets。

编码愉快！！！