猜数字游戏

Question

我创建了一个简单的数字猜谜游戏，并且尝试修复无意输入的错误。

当程序要求进行猜测时，如果用户按回车键而不输入数字，则会出现此错误：

回溯（最近通话最近）：文件 “ / Users / tom / Documents /使用Python / RandomNumberGame.py自动化”，行 13，在 guess = int（input（））ValueError：int（）以10为底的无效文字：''

我希望它打印“请输入数字”。

我是编程的新手，已经开始阅读“使用Python自动化无聊的东西”。预先感谢！

原始代码不包含

elif guess == ' ':
     print('Please enter a number')

但是目标是让程序说输入是否为空

我尝试添加：

guess = int(input()) or str(input())

没有任何进展

猜数字游戏

import random
print('Hello, What is your name?')
name = input()
print('Well, ' + name + ', I am thinking of a number between 1 and 1000, You have 10 guesses to figure it out. Good luck!')
secretNumber = random.randint(1,1000)

print('DEBUG: Secret number is ' + str(secretNumber))

for guessesTaken in range(1,11):
    print('Take a guess.')
    guess = int(input())


    if guess < secretNumber: 
          print('Your guess is too low.')
    elif guess > secretNumber:
          print('Your guess is too high.')
    elif guess == ' ':
          print('Please enter a number')
    else:
          break #This condition is for the correct guess

if guess == secretNumber:
          print('Good job ' + name + '! You guessed the number in ' + str(guessesTaken) + ' guesses!')
else:
          print('Too many guesses, The number I was thinking of was ' + str(secretNumber))

Answer 1

变量guess是整数值，因为您正在将输入转换为整数guess = int(input())。如果输入是空格或任何其他无效字符串（无法解析为数字），则会引发ValueError exception-这就是收到该错误消息的原因。

您可以使用try/except块来处理Python中何时发生特定异常。它的工作原理如下：

try:
  # execute some code
except SomeException:
  # handle the exception

将尝试执行try块中的代码，并且如果运行该代码导致引发SomeException异常，则将运行except块中的代码。

在您的特定情况下，您想要做的是处理ValueError异常，因此您可以像这样将相关代码包装在try/except中：

...
for guessesTaken in range(1,11):
    print('Take a guess.')

    try:
        guess = int(input())
        if guess < secretNumber: 
              print('Your guess is too low.')
        elif guess > secretNumber:
              print('Your guess is too high.')
        elif guess == ' ':
              print('Please enter a number')
        else:
              break #This condition is for the correct guess
    except ValueError:
        print('Please enter a number')
...

如果您希望无效输入不引起猜测，则可以像这样实现游戏：

...
guessesTaken = 1
while True:
    print('Take a guess.')

    try:
        guess = int(input())
        if guess < secretNumber: 
              print('Your guess is too low.')
        elif guess > secretNumber:
              print('Your guess is too high.')
        elif guess == ' ':
              print('Please enter a number')
        else:
              break #This condition is for the correct guess
    except ValueError:
        print('Please enter a number')
        continue # go back to the start of loop (without incrementing guessesTaken)

    guessesTaken += 1
    if guessesTaken >= 10:
        break
...

Answer 2

如果您也不想错过尝试，而用户只是按Enter键而不输入数字。您可以像这样在pure id块中使用try-except

Applicative

如果用户在尝试次数保持不变的情况下意外按下Enter键，则将允许用户继续游戏。

完整代码：

while True:

我需要帮助来解决简单的随机数字游戏中的ValueError

猜数字游戏

2 个答案: