请求用户进行输入，直到他们给出有效的响应

Question

我正在编写一个程序，接受用户的输入。

#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，则失败：

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'

我希望程序再次请求输入，而不是崩溃。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!

当输入非感官数据时，如何使程序请求有效的输入而不是崩溃？

我怎样才能拒绝像-1这样的值，这是一个有效的int，但在这个上下文中是没有意义的？

Answer 1

为什么你要做一个while True，然后跳出这个循环，而你也可以把你的需求放在while语句中，因为你只想在你有了年龄后停止？

age = None
while age is None:
    input_value = input("Please enter your age: ")
    try:
        # try and convert the string input to a number
        age = int(input_value)
    except ValueError:
        # tell the user off
        print("{input} is not a number, please enter a number only".format(input=input_value))
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.")

这将导致以下结果：

Please enter your age: *potato*
potato is not a number, please enter a number only
Please enter your age: *5*
You are not able to vote in the United States.

这将起作用，因为年龄永远不会有一个没有意义的价值，代码遵循您的“业务流程”逻辑