python - 如何发出错误而不是系统错误

This question already has answers here:

Converting String to Int using try/except in Python

(4个答案)

2年前关闭。

嘿，所以我在python3中有我的数字猜测器。但是如果用户输入字母而不是数字，我想给出一个错误。当我输入字母时它说

Traceback (most recent call last):
  File "nummerlol.py", line 14, in <module>
    guess = int(guess)
ValueError: invalid literal for int() with base 10:"

有没有办法改变这个错误？谢谢

import time
import random 
from secrets import randbelow
guesscount = 0
print("Hey whats your name")
name = input()
number = randbelow(20)
print("The Number is between 1 & 20")
while guesscount < 6:
    print("Guess.")
    guess = input()
    guess = int(guess)
    guesscount = guesscount + 1
    if guess < number:
        print("My Number is bigger")
    if guess > number:
        print("My Number is lower")
    if guess == number:
        break
if guess == number :
guesscount = str(guesscount)
print("Nice",name ,".")
print("You needed" ,guesscount ,"Tries")
if guess != number:
    number = str(number)
    print("Bruh,My Number was" ,number)`

最佳答案

您可以使用try-except循环。使用continue(如下所示)，用户可以继续猜测。如果您希望脚本以更具信息性的消息中断，您也可以将自己的异常替换为print和continue。

import time
import random 
from secrets import randbelow
guesscount = 0
print("Hey whats your name")
name = input()
number = randbelow(20)
print("The Number is between 1 & 20")
while guesscount < 6:
    print("Guess.")
    guess = input()
    try:
        guess = int(guess)
    except:
        print("That is not a number")
        continue
    guesscount = guesscount + 1
    if guess < number:
        print("My Number is bigger")
    if guess > number:
        print("My Number is lower")
    if guess == number:
        break
if guess == number :
    guesscount = str(guesscount)
print("Nice",name ,".")
print("You needed" ,guesscount ,"Tries")
if guess != number:
    number = str(number)
    print("Bruh,My Number was" ,number)

关于python - 如何发出错误而不是系统错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49575896/

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