我正在编写一个必须接受用户输入的程序。
#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.")
如果用户输入合理的数据,这将按预期工作。
C:\Python\Projects> canyouvote.py
Please enter your age: 23
You are able to vote in the United States!
但如果他们犯了一个错误,它就会崩溃:
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'
我希望它不要崩溃,而是再次尝试获取输入。这样地:
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!
我怎样才能做到这一点?如果我还想拒绝像
-1
,这是一个有效的
int
但在这种情况下,这是无稽之谈?
