Question

我是Python的新手。我正在编写一个程序，最初要求用户提供信息（姓名，年龄等），并且试图弄清楚如果用户的年龄在7岁以下，如何解决该问题。这是到目前为止的代码。

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print(f'OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
        break
    except:
            print("Please enter a number")

起初，我尝试在打印后的elif语句之后添加一个exit()语句，但是我意识到该程序不会退出，而是会循环回到第一个要求用户年龄的输入。我该如何做才能使程序终止，从而使未成年玩家无法进步？

谢谢大家！！

Answer 1

您提供的代码在执行此操作时效果很好：

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print(f'OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
        break
    except:
            print("Please enter a number")

age_func()

但是，如果将函数调用放在循环中，

while True:
  age_func()

它不起作用。我猜就是这样。您可以使用sys.exit()来解决：

import sys

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print(f'OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
        sys.exit() #replaced break with sys.exit()
    except:
            print("Please enter a number")

while True:            
  age_func()

Answer 2

try... except正在捕获exit()引发的异常。

一个更好的解决方案是不使用全部捕获的except，而是使用except ValueError，它只会捕获特定的错误。

Answer 3

请阅读整个帖子：）

杀死程序的解决方案是使用'sys.exit（）'。因此，您可以这样做：

import sys

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print(f'OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
        sys.exit() #replaced break with sys.exit()
    except:
            print("Please enter a number")

age_func()

不幸的是，由于您的try语句，这仍然不起作用。退出被视为错误，因此，当您尝试运行'sys.exit（）'时，即使您键入是，它也会发送“请输入数字”。这就是为什么指定异常类型很重要的原因，例如：

import sys

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print(f'OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
        sys.exit() #replaced break with sys.exit()
    except ValueError: #Now it will only except if what you typed wasn't a number, so it will exit properly.
            print("Please enter a number")

age_func()

希望这会有所帮助

Answer 4

不需要理解elif语句，因为它可以理解，如果if age>7为false，则表示年龄总是小于7或7。 def age_func(): while True: age = int(input('How old are you? ')) if age > 7: print(f'OK! You are {age} years old') else: print("Too young for this game. Come back in a few years.") break

Answer 5

import sys

def age_func():
  while True:
    try:
        age = int(input('How old are you? '))
        if age > 7:
            print('OK! You are {age} years old')
        elif age < 7:
            print("Too young for this game. Come back in a few years.")
            sys.exit() 
    except ValueError:
            print("Please enter a number")

age_func()

（Python 3.2）：当if语句不满足条件时退出

5 个答案: