使用自定义ValidationError和(可选的)整数输入范围验证的输入验证的另一个解决方案:

class ValidationError(ValueError): 
    """Special validation error - its message is supposed to be printed"""
    pass

def RangeValidator(text,num,r):
    """Generic validator - raises 'text' as ValidationError if 'num' not in range 'r'."""
    if num in r:
        return num
    raise ValidationError(text)

def ValidCol(c): 
    """Specialized column validator providing text and range."""
    return RangeValidator("Columns must be in the range of 0 to 3 (inclusive)", 
                          c, range(4))

def ValidRow(r): 
    """Specialized row validator providing text and range."""
    return RangeValidator("Rows must be in the range of 5 to 15(exclusive)",
                          r, range(5,15))

用法:

def GetInt(text, validator=None):
    """Aks user for integer input until a valid integer is given. If provided, 
    a 'validator' function takes the integer and either raises a 
    ValidationError to be printed or returns the valid number. 
    Non integers display a simple error message."""
    print()
    while True:
        n = input(text)
        try:
            n = int(n)

            return n if validator is None else validator(n)

        except ValueError as ve:
            # prints ValidationErrors directly - else generic message:
            if isinstance(ve, ValidationError):
                print(ve)
            else:
                print("Invalid input: ", n)


column = GetInt("Pleased enter column: ", ValidCol)
row = GetInt("Pleased enter row: ", ValidRow)
print( row, column)

输出:

Pleased enter column: 22
Columns must be in the range of 0 to 3 (inclusive)
Pleased enter column: -2
Columns must be in the range of 0 to 3 (inclusive)
Pleased enter column: 2
Pleased enter row: a
Invalid input:  a
Pleased enter row: 72
Rows must be in the range of 5 to 15(exclusive)
Pleased enter row: 9  

9, 2

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“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')

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.