猜游戏isdigit帮助python

Question

我创建了一个猜谜游戏，只允许用户键入4个数字。我的错误是“'int'对象没有属性'isdigit'”。我正在尝试这样做，以便当用户键入字母时显示错误消息，例如“仅键入数字”，并让用户再次猜测。有人可以给我建议吗？

谢谢！

对不起，我的英语（不是我的母语）

import random

n = random.randint(0, 9999)

guesses = 0

print()

while True: 

  guess = (input("Enter number from 0 to 9999"))

  guess = int(guess)

  if not guess.isdigit():

      print("Only numbers are allowed")

  else:

    guess = int(guess)

    guesses = guesses + 1

    if len(str(guess)) != 4:

      print (guesses, guess, "Invalid! 4 characters only")

      print()

    elif guess < n:

      print (guesses, guess, "too low")

      print()

    elif guess > n:

      print (guesses, guess, "too high")

      print()

    elif guess == n:

      break

print (guesses ,guess, "You guessed it!")

Answer 1

isdigit 只能为字符串调用，不能为 int您必须检查给定的字符串是否可以int通过调用isdigit转换为，然后转换为。intif not guess.isdigit():&nbsp; &nbsp; print("Only numbers are allowed")guess = int(guess)

Answer 2

Python鼓励使用EAFP，因此您可以编写以下代码：guess = input("Enter number from 0 to 9999")try:&nbsp; &nbsp; guess = int(guess)except ValueError:&nbsp; &nbsp; print("Only numbers are allowed")&nbsp; &nbsp; continue当从输入字符串到整数的转换失败时，内置转换器将引发一个ValueError。当您发现一个此类错误时，您就知道它不是有效的整数，可以采取相应的措施。