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")

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')

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.")

最简单的方法是将输入法放入while循环中。当您得到错误的输入时使用continue，当您满意时跳出循环。

当你的输入可能引发一个异常

使用try和except检测用户何时输入无法解析的数据。

while True:
    try:
        # Note: Python 2.x users should use raw_input, the equivalent of 3.x's input
        age = int(input("Please enter your age: "))
    except ValueError:
        print("Sorry, I didn't understand that.")
        #better try again... Return to the start of the loop
        continue
    else:
        #age was successfully parsed!
        #we're ready to exit the loop.
        break
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.")

实现您自己的验证规则

如果想拒绝Python可以成功解析的值，可以添加自己的验证逻辑。

while True:
    data = input("Please enter a loud message (must be all caps): ")
    if not data.isupper():
        print("Sorry, your response was not loud enough.")
        continue
    else:
        #we're happy with the value given.
        #we're ready to exit the loop.
        break

while True:
    data = input("Pick an answer from A to D:")
    if data.lower() not in ('a', 'b', 'c', 'd'):
        print("Not an appropriate choice.")
    else:
        break

结合异常处理和自定义验证

以上两种技术都可以组合成一个循环。

while True:
    try:
        age = int(input("Please enter your age: "))
    except ValueError:
        print("Sorry, I didn't understand that.")
        continue

    if age < 0:
        print("Sorry, your response must not be negative.")
        continue
    else:
        #age was successfully parsed, and we're happy with its value.
        #we're ready to exit the loop.
        break
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.")

把它封装在一个函数中

如果您需要向用户询问许多不同的值，那么将这些代码放在函数中可能会很有用，这样您就不必每次都重新键入。

def get_non_negative_int(prompt):
    while True:
        try:
            value = int(input(prompt))
        except ValueError:
            print("Sorry, I didn't understand that.")
            continue

        if value < 0:
            print("Sorry, your response must not be negative.")
            continue
        else:
            break
    return value

age = get_non_negative_int("Please enter your age: ")
kids = get_non_negative_int("Please enter the number of children you have: ")
salary = get_non_negative_int("Please enter your yearly earnings, in dollars: ")

把它们放在一起

你可以扩展这个想法来创建一个非常通用的输入函数:

def sanitised_input(prompt, type_=None, min_=None, max_=None, range_=None):
    if min_ is not None and max_ is not None and max_ < min_:
        raise ValueError("min_ must be less than or equal to max_.")
    while True:
        ui = input(prompt)
        if type_ is not None:
            try:
                ui = type_(ui)
            except ValueError:
                print("Input type must be {0}.".format(type_.__name__))
                continue
        if max_ is not None and ui > max_:
            print("Input must be less than or equal to {0}.".format(max_))
        elif min_ is not None and ui < min_:
            print("Input must be greater than or equal to {0}.".format(min_))
        elif range_ is not None and ui not in range_:
            if isinstance(range_, range):
                template = "Input must be between {0.start} and {0.stop}."
                print(template.format(range_))
            else:
                template = "Input must be {0}."
                if len(range_) == 1:
                    print(template.format(*range_))
                else:
                    expected = " or ".join((
                        ", ".join(str(x) for x in range_[:-1]),
                        str(range_[-1])
                    ))
                    print(template.format(expected))
        else:
            return ui

用法如下:

age = sanitised_input("Enter your age: ", int, 1, 101)
answer = sanitised_input("Enter your answer: ", str.lower, range_=('a', 'b', 'c', 'd'))

常见的陷阱，以及为什么你应该避免它们

冗余输入语句的冗余使用

这种方法有效，但通常被认为是糟糕的风格:

data = input("Please enter a loud message (must be all caps): ")
while not data.isupper():
    print("Sorry, your response was not loud enough.")
    data = input("Please enter a loud message (must be all caps): ")

最初它可能看起来很吸引人，因为它比while True方法更短，但它违反了软件开发的“不要重复自己”原则。这增加了系统中出现错误的可能性。如果您想通过将输入更改为raw_input来向后移植到2.7，但不小心只更改了上面的第一个输入，该怎么办?这是一个等待发生的SyntaxError。

递归会破坏你的堆栈

如果您刚刚学习过递归，您可能会在get_non_negative_int中使用它，这样就可以处理while循环。

def get_non_negative_int(prompt):
    try:
        value = int(input(prompt))
    except ValueError:
        print("Sorry, I didn't understand that.")
        return get_non_negative_int(prompt)

    if value < 0:
        print("Sorry, your response must not be negative.")
        return get_non_negative_int(prompt)
    else:
        return value

这在大多数情况下似乎工作正常，但如果用户输入无效数据的次数足够多，脚本将以RuntimeError终止:超出最大递归深度。你可能认为“没有傻瓜会连续犯1000个错误”，但你低估了傻瓜的聪明才智!

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。

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。