ValueError: int() 以 10 为底的无效文字:'stop'
- 作者: 南角先生
- 来源: 51数据库
- 2023-01-30
问题描述
每次我尝试编写代码时它都可以工作,但是当我输入 'stop' 时,它会给我一个错误:
Every time I try me code it works but when I type in 'stop' it gives me an error:
ValueError: int() 以 10 为底的无效文字:'stop'
ValueError: invalid literal for int() with base 10: 'stop'
def guessingGame():
global randomNum
guessTry = 3
while True:
guess = input('Guess a Number between 1 - 10, You have 3 Tries, or Enter Stop: ')
if int(guess) == randomNum:
print('Correct')
break
if int(guess) < randomNum:
print('Too Low')
guessTry = guessTry - 1
print('You have, ' + str(guessTry) + ' Guesses Left')
if int(guess) > randomNum:
print('Too High')
guessTry = guessTry - 1
print('You have, ' + str(guessTry) + ' Guesses Left')
if guessTry == 0:
print('You have no more tries')
return
if str(guess) == 'stop' or str(guess) == 'Stop':
break
推荐答案
传递给 int() 的字符串应该只包含数字:
The string passed to int() should only contain digits:
>>> int("stop")
Traceback (most recent call last):
File "<ipython-input-114-e5503af2dc1c>", line 1, in <module>
int("stop")
ValueError: invalid literal for int() with base 10: 'stop'
快速解决方法是在此处使用 异常处理:
A quick fix will be to use exception handling here:
def guessingGame():
global randomNum
global userScore
guessTry = 3
while True:
guess = input('Guess a Number between 1 - 10, You have 3 Tries, or Enter Stop: ')
try:
if int(guess) == randomNum:
print('Correct')
break
if int(guess) < randomNum:
print('Too Low')
guessTry = guessTry - 1
print('You have, ' + str(guessTry) + ' Guesses Left')
if int(guess) > randomNum:
print('Too High')
guessTry = guessTry - 1
print('You have, ' + str(guessTry) + ' Guesses Left')
if guessTry == 0:
print('You have no more tries')
return
except ValueError:
#no need of str() here
if guess.lower() == 'stop':
break
guessingGame()
您可以使用 guess.lower() == 'stop' 来匹配stop"的任何大小写组合:
And you can use guess.lower() == 'stop' to match any uppercase-lowercase combination of "stop":
>>> "Stop".lower() == "stop" True >>> "SToP".lower() == "stop" True >>> "sTOp".lower() == "stop" True
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