我正在编写一个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.")
只要用户输入有意义的数据,程序就能正常工作。
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:
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。是这样的:
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!
我如何要求有效输入而不是崩溃或接受无效值(例如-1)?
下面的代码可能会有所帮助。
age=(lambda i,f: f(i,f))(input("Please enter your age: "),lambda i,f: i if i.isdigit() else f(input("Please enter your age: "),f))
print("You are able to vote in the united states" if int(age)>=18 else "You are not able to vote in the united states",end='')
如果你想要最大尝试次数,比如3次,请使用下面的代码
age=(lambda i,n,f: f(i,n,f))(input("Please enter your age: "),1,lambda i,n,f: i if i.isdigit() else (None if n==3 else f(input("Please enter your age: "),n+1,f)))
print("You are able to vote in the united states" if age and int(age)>=18 else "You are not able to vote in the united states",end='')
注意:这里使用递归。
使用递归函数的持久用户输入:
字符串
def askName():
return input("Write your name: ").strip() or askName()
name = askName()
整数
def askAge():
try: return int(input("Enter your age: "))
except ValueError: return askAge()
age = askAge()
最后,问题要求:
def askAge():
try: return int(input("Enter your age: "))
except ValueError: return askAge()
age = askAge()
responseAge = [
"You are able to vote in the United States!",
"You are not able to vote in the United States.",
][int(age < 18)]
print(responseAge)