def validate_age(age):
    if age >=0 :
        return True
    return False

while True:
    try:
        age = int(raw_input("Please enter your age:"))
        if validate_age(age): break
    except ValueError:
        print "Error: Invalid age."

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.

所以，我最近在搞一些类似的事情，我想出了下面的解决方案，它使用一种方法来获取输入，拒绝垃圾，甚至在它以任何逻辑方式检查之前。

Read_single_keypress()礼貌https://stackoverflow.com/a/6599441/4532996

def read_single_keypress() -> str:
    """Waits for a single keypress on stdin.
    -- from :: https://stackoverflow.com/a/6599441/4532996
    """

    import termios, fcntl, sys, os
    fd = sys.stdin.fileno()
    # save old state
    flags_save = fcntl.fcntl(fd, fcntl.F_GETFL)
    attrs_save = termios.tcgetattr(fd)
    # make raw - the way to do this comes from the termios(3) man page.
    attrs = list(attrs_save) # copy the stored version to update
    # iflag
    attrs[0] &= ~(termios.IGNBRK | termios.BRKINT | termios.PARMRK
                  | termios.ISTRIP | termios.INLCR | termios. IGNCR
                  | termios.ICRNL | termios.IXON )
    # oflag
    attrs[1] &= ~termios.OPOST
    # cflag
    attrs[2] &= ~(termios.CSIZE | termios. PARENB)
    attrs[2] |= termios.CS8
    # lflag
    attrs[3] &= ~(termios.ECHONL | termios.ECHO | termios.ICANON
                  | termios.ISIG | termios.IEXTEN)
    termios.tcsetattr(fd, termios.TCSANOW, attrs)
    # turn off non-blocking
    fcntl.fcntl(fd, fcntl.F_SETFL, flags_save & ~os.O_NONBLOCK)
    # read a single keystroke
    try:
        ret = sys.stdin.read(1) # returns a single character
    except KeyboardInterrupt:
        ret = 0
    finally:
        # restore old state
        termios.tcsetattr(fd, termios.TCSAFLUSH, attrs_save)
        fcntl.fcntl(fd, fcntl.F_SETFL, flags_save)
    return ret

def until_not_multi(chars) -> str:
    """read stdin until !(chars)"""
    import sys
    chars = list(chars)
    y = ""
    sys.stdout.flush()
    while True:
        i = read_single_keypress()
        _ = sys.stdout.write(i)
        sys.stdout.flush()
        if i not in chars:
            break
        y += i
    return y

def _can_you_vote() -> str:
    """a practical example:
    test if a user can vote based purely on keypresses"""
    print("can you vote? age : ", end="")
    x = int("0" + until_not_multi("0123456789"))
    if not x:
        print("\nsorry, age can only consist of digits.")
        return
    print("your age is", x, "\nYou can vote!" if x >= 18 else "Sorry! you can't vote")

_can_you_vote()

您可以在这里找到完整的模块。

例子:

$ ./input_constrain.py
can you vote? age : a
sorry, age can only consist of digits.
$ ./input_constrain.py 
can you vote? age : 23<RETURN>
your age is 23
You can vote!
$ _

注意，这个实现的本质是，一旦读取非数字的内容，它就会关闭stdin。我没有在a后面按回车键，但我需要在数字后面按。

您可以将此与同一模块中的thismany()函数合并，以只允许(比如说)三位数字。

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.")

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')

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。