Question

在下面的程序中，如果我按下回车键，我会得到：

Traceback (most recent call last):
  File "C:/Users/Claude/Desktop/practice.py", line 11, in <module>
    guess = int(input("Your guess: "))
ValueError: invalid literal for int() with base 10: ''*

在寻找方法来检测＆＃34;输入＆＃34;时，我找不到答案。钥匙被压了。有人可以帮忙吗？

from random import randint

# Generates a number from 1 through 10 inclusive
random_number = randint(1, 10)

guesses_left = 6

while (guesses_left > 0):

        guess = int(input("Your guess: "))
        if (guess == random_number):
            print ("You win!")
            ans = input("Do you want to play again?:")
        if (ans == 'y' or ans == 'Y'):
            guesses_left = 5
            guess = int(input("Your guess: "))
        else:
            break
    #elif (guess == ""):
    #   print ("Please enter a number between 1 and 10")
    elif(guess != random_number):        
        guesses_left -= 1
else:
    print ("You lose!")

Answer 1

当您在input提示符下按Enter键时，您基本上输入了一个空字符串。所以你会尝试将一个空字符串转换为一个不起作用的数字。相反，您应该首先查看字符串并查看它是否为空，在这种情况下用户只需按Enter键，否则尝试转换它：

guess = input("Your guess: ")
if not guess:
    print("You didn't enter anything. So let's abort.")
    break

try:
    # try converting it to a number
    guess = int(guess)
except ValueError:
    # ValueError is raised when that didn't work
    print("That wasn't a number!")
else:
    # otherwise we now have a number which we can use
    if guess == random_number:
        print('You win')

    # …

Answer 2

因为Your guess！

int(x)函数将数字或字符串x转换为整数，如果没有给出参数，则返回0。如果x是数字，则它可以是普通整数，长整数或浮点数。如果x是浮点，则转换将截断为零。如果参数超出整数范围，则函数返回一个long对象。如果x不是数字或者给定了base，那么x必须是一个字符串或Unicode对象，表示以radix为基数的整数文字。！

如果你有python 2.x，你可以使用raw_input()函数来获取用户的一般输入。

Answer 3

我通过使用ord（）获得“输入”代码来解决我的问题。打印（ord（key.char）） 13 如果ord（key.char）== 13：打印（“ Enter Pressed”）其他：打印（“按下任何其他键”）

如何检查在Python中是否按下了Enter键？

3 个答案: