这篇文章主要介绍如何解决python面对用户无意义输入的问题，文中介绍的非常详细，具有一定的参考价值，感兴趣的小伙伴们一定要看完！
问题
正在编写一个接受用户输入的程序。
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#note: Python 2.7 users should use `raw_input`, the equivalent of 3.X's `input`
age = int(input("Please enter your age: "))
if age >= 18:
print("You are able to vote in the United States!")
else:
print("You are not able to vote in the United States.")只要用户输入有意义的数据，程序就会按预期工作。
C:\Python\Projects> canyouvote.py
Please enter your age: 23
You are able to vote in the United States!但如果用户输入无效数据，它就会失败：
C:\Python\Projects> canyouvote.py
Please enter your age: dickety six
Traceback (most recent call last):
File "canyouvote.py", line 1, in <module>
age = int(input("Please enter your age: "))
ValueError: invalid literal for int() with base 10: 'dickety six'而不是崩溃，我希望程序再次要求输入。像这样：
C:\Python\Projects> canyouvote.py
Please enter your age: dickety six
Sorry, I didn't understand that.
Please enter your age: 26
You are able to vote in the United States!当输入无意义的数据时，如何让程序要求有效输入而不是崩溃？
我如何拒绝像 那样的值-1，int在这种情况下这是一个有效但无意义的值？
解决方法
完成此操作的最简单方法是将input方法放入 while 循环中。使用continue时，你会得到错误的输入，并break退出循环。
当您的输入可能引发异常时
使用try和except检测用户何时输入无法解析的数据。
while True:
try:
# Note: Python 2.x users should use raw_input, the equivalent of 3.x's input
age = int(input("Please enter your age: "))
except ValueError:
print("Sorry, I didn't understand that.")
#better try again... Return to the start of the loop
continue
else:
#age was successfully parsed!
#we're ready to exit the loop.
break
if age >= 18:
print("You are able to vote in the United States!")
else:
print("You are not able to vote in the United States.")实现你自己的验证规则
如果要拒绝 Python 可以成功解析的值，可以添加自己的验证逻辑。
while True:
data = input("Please enter a loud message (must be all caps): ")
if not data.isupper():
print("Sorry, your response was not loud enough.")
continue
else:
#we're happy with the value given.
#we're ready to exit the loop.
break
while True:
data = input("Pick an answer from A to D:")
if data.lower() not in ('a', 'b', 'c', 'd'):
print("Not an appropriate choice.")
else:
Break结合异常处理和自定义验证
以上两种技术都可以组合成一个循环。
while True:
try:
age = int(input("Please enter your age: "))
except ValueError:
print("Sorry, I didn't understand that.")
continue
if age < 0:
print("Sorry, your response must not be negative.")
continue
else:
#age was successfully parsed, and we're happy with its value.
#we're ready to exit the loop.
break
if age >= 18:
print("You are able to vote in the United States!")
else:
print("You are not able to vote in the United States.")以上是“如何解决python面对用户无意义输入的问题”这篇文章的所有内容，感谢各位的阅读！希望分享的内容对大家有帮助，更多相关知识，欢迎关注亿速云行业资讯频道！
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原文链接：https://www.py.cn/jishu/jichu/32299.html
