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."

使用递归函数的持久用户输入:

字符串

def askName():
    return input("Write your name: ").strip() or askName()

name = askName()

整数

def askAge():
    try: return int(input("Enter your age: "))
    except ValueError: return askAge()

age = askAge()

最后，问题要求:

def askAge():
    try: return int(input("Enter your age: "))
    except ValueError: return askAge()

age = askAge()

responseAge = [
    "You are able to vote in the United States!",
    "You are not able to vote in the United States.",
][int(age < 18)]

print(responseAge)

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.") 

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.

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

函数方法或“看，妈妈，没有循环!”：

from itertools import chain, repeat

prompts = chain(["Enter a number: "], repeat("Not a number! Try again: "))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies))
print(valid_response)

Enter a number:  a
Not a number! Try again:  b
Not a number! Try again:  1
1

或者如果你想有一个“错误输入”的消息从输入提示符中分离出来，就像在其他答案中:

prompt_msg = "Enter a number: "
bad_input_msg = "Sorry, I didn't understand that."
prompts = chain([prompt_msg], repeat('\n'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies))
print(valid_response)

Enter a number:  a
Sorry, I didn't understand that.
Enter a number:  b
Sorry, I didn't understand that.
Enter a number:  1
1

它是如何工作的?

prompts = chain(["Enter a number: "], repeat("Not a number! Try again: ")) This combination of itertools.chain and itertools.repeat will create an iterator which will yield strings "Enter a number: " once, and "Not a number! Try again: " an infinite number of times: for prompt in prompts: print(prompt) Enter a number: Not a number! Try again: Not a number! Try again: Not a number! Try again: # ... and so on replies = map(input, prompts) - here map will apply all the prompts strings from the previous step to the input function. E.g.: for reply in replies: print(reply) Enter a number: a a Not a number! Try again: 1 1 Not a number! Try again: it doesn't care now it doesn't care now # and so on... We use filter and str.isdigit to filter out those strings that contain only digits: only_digits = filter(str.isdigit, replies) for reply in only_digits: print(reply) Enter a number: a Not a number! Try again: 1 1 Not a number! Try again: 2 2 Not a number! Try again: b Not a number! Try again: # and so on... And to get only the first digits-only string we use next.

其他验证规则:

String methods: Of course you can use other string methods like str.isalpha to get only alphabetic strings, or str.isupper to get only uppercase. See docs for the full list. Membership testing: There are several different ways to perform it. One of them is by using __contains__ method: from itertools import chain, repeat fruits = {'apple', 'orange', 'peach'} prompts = chain(["Enter a fruit: "], repeat("I don't know this one! Try again: ")) replies = map(input, prompts) valid_response = next(filter(fruits.__contains__, replies)) print(valid_response) Enter a fruit: 1 I don't know this one! Try again: foo I don't know this one! Try again: apple apple Numbers comparison: There are useful comparison methods which we can use here. For example, for __lt__ (<): from itertools import chain, repeat prompts = chain(["Enter a positive number:"], repeat("I need a positive number! Try again:")) replies = map(input, prompts) numeric_strings = filter(str.isnumeric, replies) numbers = map(float, numeric_strings) is_positive = (0.).__lt__ valid_response = next(filter(is_positive, numbers)) print(valid_response) Enter a positive number: a I need a positive number! Try again: -5 I need a positive number! Try again: 0 I need a positive number! Try again: 5 5.0 Or, if you don't like using dunder methods (dunder = double-underscore), you can always define your own function, or use the ones from the operator module. Path existance: Here one can use pathlib library and its Path.exists method: from itertools import chain, repeat from pathlib import Path prompts = chain(["Enter a path: "], repeat("This path doesn't exist! Try again: ")) replies = map(input, prompts) paths = map(Path, replies) valid_response = next(filter(Path.exists, paths)) print(valid_response) Enter a path: a b c This path doesn't exist! Try again: 1 This path doesn't exist! Try again: existing_file.txt existing_file.txt

限制尝试次数:

如果您不想通过无数次地询问用户某个问题来折磨用户，您可以在itertools.repeat调用中指定一个限制。这可以与为下一个函数提供默认值相结合:

from itertools import chain, repeat

prompts = chain(["Enter a number:"], repeat("Not a number! Try again:", 2))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies), None)
print("You've failed miserably!" if valid_response is None else 'Well done!')

Enter a number: a
Not a number! Try again: b
Not a number! Try again: c
You've failed miserably!

预处理输入数据:

有时，如果用户不小心以大写形式提供输入，或者在字符串的开头或结尾使用空格，我们不想拒绝输入。为了考虑这些简单的错误，我们可以通过应用str.lower和str.strip方法对输入数据进行预处理。例如，在成员测试的情况下，代码看起来像这样:

from itertools import chain, repeat

fruits = {'apple', 'orange', 'peach'}
prompts = chain(["Enter a fruit: "], repeat("I don't know this one! Try again: "))
replies = map(input, prompts)
lowercased_replies = map(str.lower, replies)
stripped_replies = map(str.strip, lowercased_replies)
valid_response = next(filter(fruits.__contains__, stripped_replies))
print(valid_response)

Enter a fruit:  duck
I don't know this one! Try again:     Orange
orange

在有许多函数用于预处理的情况下，使用一个函数执行函数组合可能更容易。例如，使用这里的一个:

from itertools import chain, repeat

from lz.functional import compose

fruits = {'apple', 'orange', 'peach'}
prompts = chain(["Enter a fruit: "], repeat("I don't know this one! Try again: "))
replies = map(input, prompts)
process = compose(str.strip, str.lower)  # you can add more functions here
processed_replies = map(process, replies)
valid_response = next(filter(fruits.__contains__, processed_replies))
print(valid_response)

Enter a fruit:  potato
I don't know this one! Try again:   PEACH
peach

组合验证规则:

例如，对于一个简单的情况，当程序要求年龄在1到120之间时，可以添加另一个过滤器:

from itertools import chain, repeat

prompt_msg = "Enter your age (1-120): "
bad_input_msg = "Wrong input."
prompts = chain([prompt_msg], repeat('\n'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
numeric_replies = filter(str.isdigit, replies)
ages = map(int, numeric_replies)
positive_ages = filter((0).__lt__, ages)
not_too_big_ages = filter((120).__ge__, positive_ages)
valid_response = next(not_too_big_ages)
print(valid_response)

但是在有很多规则的情况下，最好实现一个执行逻辑连接的函数。在下面的例子中，我将使用一个现成的例子:

from functools import partial
from itertools import chain, repeat

from lz.logical import conjoin


def is_one_letter(string: str) -> bool:
    return len(string) == 1


rules = [str.isalpha, str.isupper, is_one_letter, 'C'.__le__, 'P'.__ge__]

prompt_msg = "Enter a letter (C-P): "
bad_input_msg = "Wrong input."
prompts = chain([prompt_msg], repeat('\n'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
valid_response = next(filter(conjoin(*rules), replies))
print(valid_response)

Enter a letter (C-P):  5
Wrong input.
Enter a letter (C-P):  f
Wrong input.
Enter a letter (C-P):  CDE
Wrong input.
Enter a letter (C-P):  Q
Wrong input.
Enter a letter (C-P):  N
N

不幸的是，如果有人需要为每个失败的情况定制消息，那么恐怕没有非常实用的方法。或者，至少我找不到。