虽然try/except块可以工作，但是使用str.isdigit()可以更快更清晰地完成此任务。

while True:
    age = input("Please enter your age: ")
    if age.isdigit():
        age = int(age)
        break
    else:
        print("Invalid number '{age}'. Try again.".format(age=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.")

为什么你要做一个while True，然后跳出这个循环，而你也可以把你的要求放在while语句中因为你想要的是一旦你有了年龄就停止?

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.

这是可行的，因为年龄永远不会有一个没有意义的值，代码遵循“业务流程”的逻辑。

好问题!您可以尝试以下代码。=)

这段代码使用ast.literal_eval()来查找输入的数据类型(age)。然后按照以下算法:

请用户输入年龄。 1.1. 如果age为float或int数据类型: 检查年龄>是否=18。如果age>=18，打印相应的输出并退出。 检查0<年龄<18。如果0<age<18，打印适当的输出并退出。 如果age<=0，请用户再次输入age的有效数字(即返回步骤1)。 1.2. 如果age不是float或int数据类型，则要求用户再次输入她/他的年龄(即返回第1步)。

这是代码。

from ast import literal_eval

''' This function is used to identify the data type of input data.'''
def input_type(input_data):
    try:
        return type(literal_eval(input_data))
    except (ValueError, SyntaxError):
        return str

flag = True

while(flag):
    age = raw_input("Please enter your age: ")

    if input_type(age)==float or input_type(age)==int:
        if eval(age)>=18: 
            print("You are able to vote in the United States!") 
            flag = False 
        elif eval(age)>0 and eval(age)<18: 
            print("You are not able to vote in the United States.") 
            flag = False
        else: print("Please enter a valid number as your age.")

    else: print("Sorry, I didn't understand that.") 

使用try-except来处理错误并重复一次:

while True:
    try:
        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.")
    except Exception as e:
        print("please enter number")

虽然公认的答案是惊人的。我也想分享一个快速解决这个问题的方法。(这也解决了负年龄问题。)

f=lambda age: (age.isdigit() and ((int(age)>=18  and "Can vote" ) or "Cannot vote")) or \
f(input("invalid input. Try again\nPlease enter your age: "))
print(f(input("Please enter your age: ")))

附注:此代码用于python 3.x。

您总是可以应用简单的if-else逻辑，并在for循环的同时向代码中添加一个if逻辑。

while True:
     age = int(input("Please enter your age: "))
     if (age >= 18)  : 
         print("You are able to vote in the United States!")
     if (age < 18) & (age > 0):
         print("You are not able to vote in the United States.")
     else:
         print("Wrong characters, the input must be numeric")
         continue

这将是一个无限的厕所，你将被要求进入一个无限的时代。