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。

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。

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。

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")

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.")