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.")

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。

使用“while”语句，直到用户输入一个真值，如果输入值不是一个数字或它是一个空值跳过它，并尝试再次询问，等等。 举例来说，我试图真正地回答你的问题。如果我们假设我们的年龄在1到150之间，那么输入值被接受，否则它是一个错误的值。 对于终止程序，用户可以使用0键并输入它作为一个值。

注意:阅读代码顶部的注释。

# If your input value is only a number then use "Value.isdigit() == False".
# If you need an input that is a text, you should remove "Value.isdigit() == False".
def Input(Message):
    Value = None
    while Value == None or Value.isdigit() == False:
        try:        
            Value = str(input(Message)).strip()
        except Exception:
            Value = None
    return Value

# Example:
age = 0
# If we suppose that our age is between 1 and 150 then input value accepted,
# else it's a wrong value.
while age <=0 or age >150:
    age = int(Input("Please enter your age: "))
    # For terminating program, the user can use 0 key and enter it as an a value.
    if age == 0:
        print("Terminating ...")
        exit(0)
        
if age >= 18 and age <=150: 
    print("You are able to vote in the United States!")
else:
    print("You are not able to vote in the United States.")

基于Daniel Q和Patrick Artner的优秀建议， 这里有一个更普遍的解决方案。

# Assuming Python3
import sys

class ValidationError(ValueError):  # thanks Patrick Artner
    pass

def validate_input(prompt, cast=str, cond=(lambda x: True), onerror=None):
    if onerror==None: onerror = {}
    while True:
        try:
            data = cast(input(prompt))
            if not cond(data): raise ValidationError
            return data
        except tuple(onerror.keys()) as e:  # thanks Daniel Q
            print(onerror[type(e)], file=sys.stderr)

我选择了显式的if和raise语句，而不是assert， 因为断言检查可能被关闭， 而验证应始终开启以提供健壮性。

这可以用来获得不同种类的输入， 使用不同的验证条件。 例如:

# No validation, equivalent to simple input:
anystr = validate_input("Enter any string: ")

# Get a string containing only letters:
letters = validate_input("Enter letters: ",
    cond=str.isalpha,
    onerror={ValidationError: "Only letters, please!"})

# Get a float in [0, 100]:
percentage = validate_input("Percentage? ",
    cast=float, cond=lambda x: 0.0<=x<=100.0,
    onerror={ValidationError: "Must be between 0 and 100!",
             ValueError: "Not a number!"})

或者，回答最初的问题:

age = validate_input("Please enter your age: ",
        cast=int, cond=lambda a:0<=a<150,
        onerror={ValidationError: "Enter a plausible age, please!",
                 ValueError: "Enter an integer, please!"})
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.")

我是Unix哲学“只做一件事并把它做好”的忠实粉丝。捕获用户输入并验证它是两个独立的步骤:

使用get_input提示用户输入，直到输入成功 使用可以传递给get_input的验证器函数进行验证

它可以保持简单如(Python 3.8+，使用walrus操作符):

def get_input(
    prompt="Enter a value: ",
    validator=lambda x: True,
    error_message="Invalid input. Please try again.",
):
    while not validator(value := input(prompt)):
        print(error_message)
    return value

def is_positive_int(value):
    try:
        return int(value) >= 0
    except ValueError:
        return False

if __name__ == "__main__":
    val = get_input("Give a positive number: ", is_positive_int)
    print(f"OK, thanks for {val}")

示例运行:

Give a positive number: -5
Invalid input. Please try again.
Give a positive number: asdf
Invalid input. Please try again.
Give a positive number:
Invalid input. Please try again.
Give a positive number: 42
OK, thanks for 42

在Python < 3.8中，你可以像这样使用get_input:

def get_input(
    prompt="Enter a value: ",
    validator=lambda x: True,
    error_message="Invalid input. Please try again.",
):
    while True:
        value = input(prompt)
        if validator(value):
            return value
        print(error_message)

您还可以在终止应用程序之前处理KeyboardInterrupt并打印友好的退出消息。如果需要，可以使用计数器限制允许的重试次数。

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。