您可以尝试将其转换为整数，但如果不成功，请用户重复。

while True:
    age = input('Please enter your age: ')
    try:
        age_int = int(age)
        if age_int >= 18:
            print('You can vote in the United States!')
        else:
            print('You cannot vote in the United States.')
        break
    except:
        print('Please enter a meaningful answer.')
        

只要用户没有输入有意义的答案，while循环就会运行，但如果有意义就会中断。

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.")

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")

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。

试试这个:-

def takeInput(required):
  print 'ooo or OOO to exit'
  ans = raw_input('Enter: ')

  if not ans:
      print "You entered nothing...!"
      return takeInput(required) 

      ##  FOR Exit  ## 
  elif ans in ['ooo', 'OOO']:
    print "Closing instance."
    exit()

  else:
    if ans.isdigit():
      current = 'int'
    elif set('[~!@#$%^&*()_+{}":/\']+$').intersection(ans):
      current = 'other'
    elif isinstance(ans,basestring):
      current = 'str'        
    else:
      current = 'none'

  if required == current :
    return ans
  else:
    return takeInput(required)

## pass the value in which type you want [str/int/special character(as other )]
print "input: ", takeInput('str')

您可以编写更通用的逻辑，以允许用户只输入特定次数，因为在许多实际应用程序中都会出现相同的用例。

def getValidInt(iMaxAttemps = None):
  iCount = 0
  while True:
    # exit when maximum attempt limit has expired
    if iCount != None and iCount > iMaxAttemps:
       return 0     # return as default value

    i = raw_input("Enter no")
    try:
       i = int(i)
    except ValueError as e:
       print "Enter valid int value"
    else:
       break

    return i

age = getValidInt()
# do whatever you want to do.