使用Click：

Question

我正在编写一个必须接受用户输入的程序。

#note: Python 2.7 users should use `raw_input`, the equivalent of 3.X's `input`
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.")

如果用户输入合理数据，这将按预期工作。

C:\Python\Projects> canyouvote.py
Please enter your age: 23
You are able to vote in the United States!

但如果他们犯了错误，那就崩溃了：

C:\Python\Projects> canyouvote.py
Please enter your age: dickety six
Traceback (most recent call last):
  File "canyouvote.py", line 1, in <module>
    age = int(input("Please enter your age: "))
ValueError: invalid literal for int() with base 10: 'dickety six'

而不是崩溃，我希望它再次尝试获取输入。像这样：

C:\Python\Projects> canyouvote.py
Please enter your age: dickety six
Sorry, I didn't understand that.
Please enter your age: 26
You are able to vote in the United States!

我怎样才能做到这一点？如果我还想拒绝像-1这样的值，这是有效的int，但在此上下文中是无意义的，会怎么样？

Answer 1

实现此目的的最简单方法是将input方法放入while循环中。输入错误时使用continue，当您满意时使用break。

当您的输入可能会引发异常时

使用try and catch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.")

实施您自己的验证规则

如果要拒绝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

上述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.")

将其全部封装在函数中

如果您需要向用户询问许多不同的值，将此代码放在一个函数中可能会很有用，因此您不必每次都重新键入它。

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: ")

全部放在一起

你可以扩展这个想法，使其成为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:
                    print(template.format(" or ".join((", ".join(map(str,
                                                                     range_[:-1])),
                                                       str(range_[-1])))))
        else:
            return ui

使用如下：

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，以及为什么要避免它们

冗余使用冗余`input`语句

这种方法有效但通常被认为是不好的风格：

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): ")

它最初看起来很有吸引力，因为它比while True方法短，但它违反了Don't Repeat Yourself软件开发原则。这会增加系统中出现错误的可能性。如果您想通过将input更改为raw_input来向后移植到2.7，但却意外更改了上面的第一个input，该怎么办？它只是等待发生SyntaxError。

递归会打击你的堆栈

如果您刚刚了解了递归，那么您可能很想在get_non_negative_int中使用它，因此您可以处理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

这似乎在大多数情况下都能正常工作，但如果用户输入的数据足够多次，脚本将以RuntimeError: maximum recursion depth exceeded终止。你可能会认为＆＃34;没有傻瓜会连续犯下1000个错误＆＃34;但你却低估了傻瓜的聪明才智！

Answer 2

为什么你会做while True然后突破这个循环，同时你也可以把你的要求放在while语句中，因为你想要的就是在你有了年龄后停止？

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.")

这将产生以下结果：

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.

这将有效，因为年龄永远不会有一个没有意义的价值，而且代码遵循“业务流程”的逻辑

Answer 3

虽然接受的答案很惊人。我还想分享这个问题的快速入侵。（这也解决了负面年龄问题。）

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: ")))

P.S。此代码适用于python 3.x。

Answer 4

所以，我最近搞乱了类似的东西，我想出了以下解决方案，它使用一种获取输入的方法来拒绝垃圾，甚至在它以任何逻辑方式检查之前。< / p>

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()

您可以找到完整的模块here。

示例：

$ ./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!
$ _

请注意，此实现的性质是，只要不读取数字，就会关闭stdin。我在a之后没有进入，但我需要在数字之后。

您可以将其与同一模块中的thismany()功能合并为仅允许三位数。

Answer 5

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."

Answer 6

基于Daniel Q和Patrick Artner的出色建议，这是一个更通用的解决方案。

# Assuming Python3
import sys

class ValidationError(ValueError):  # thanks Patrick Artner
    pass

def validate_input(prompt, cast=str, cond=(lambda x: True), onerror=None):
    if onerror==None: onerror = {}
    while True:
        try:
            data = cast(input(prompt))
            if not cond(data): raise ValidationError
            return data
        except tuple(onerror.keys()) as e:  # thanks Daniel Q
            print(onerror[type(e)], file=sys.stderr)

我选择了明确的if和raise语句，而不是assert，因为断言检查可能已关闭，而应该始终启用验证以提供鲁棒性。

这可以用来获取不同种类的输入，具有不同的验证条件。例如：

# No validation, equivalent to simple input:
anystr = validate_input("Enter any string: ")

# Get a string containing only letters:
letters = validate_input("Enter letters: ",
    cond=str.isalpha,
    onerror={ValidationError: "Only letters, please!"})

# Get a float in [0, 100]:
percentage = validate_input("Percentage? ",
    cast=float, cond=lambda x: 0.0<=x<=100.0,
    onerror={ValidationError: "Must be between 0 and 100!",
             ValueError: "Not a number!"})

或者，回答原始问题：

age = validate_input("Please enter your age: ",
        cast=int, cond=lambda a:0<=a<150,
        onerror={ValidationError: "Enter a plausible age, please!",
                 ValueError: "Enter an integer, please!"})
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.")

Answer 7

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')

Answer 8

是的，我距离?晚了6年，但是这个问题值得我们提供最新的答案。

关注点分离

我是Unix哲学的忠实拥护者“做一件事情，做好事” 。在这种类型的问题中，更好的做法是将问题分解为

使用get_input询问输入，直到输入正常为止。
验证validator函数。您可以为不同的输入查询编写不同的验证器。

要求输入

它可以像（Python 3+）一样简单

def myvalidator(value):
    try:
        value = int(value)
    except ValueError:
        return False
    return value >= 0

def get_input(prompt, validator, on_validationerror):
    while True:
        value = input(prompt)
        if validator(value):
            return value
        print(on_validationerror)

示例

In [2]: get_input('Give a positive number: ', myvalidator, 'Please, try again')
Give a positive number: foobar
Please, try again
Give a positive number: -10
Please, try again
Give a positive number: 42
Out[2]: '42'

Python 3.8+注意事项

在Python 3.8+中，您可以使用海象运算符

def get_input(prompt, validator, on_validationerror):
    while not validator(value := input(prompt)):
        print(on_validationerror)
    return value

Answer 9

简单的解决方案是：

while True:
    age = int(input("Please enter your age: "))

    if (age<=0) or (age>120):
        print('Sorry, I did not understand that.Please try again')
        continue
    else:

        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.")
        break

以上代码的说明： 为了确定有效年龄，该年龄应该是正数，并且不应超过正常的生理年龄，例如，最大年龄为120。

然后我们可以询问用户年龄，如果年龄输入为负数或大于120，我们将其视为无效输入，然后要求用户重试。

输入有效输入后，我们将检查（使用嵌套的if-else语句）年龄是否大于等于18，反之亦然，并显示一条消息，表明用户是否有投票权

Answer 10

使用Click：

Click 是用于命令行界面的库，它提供了向用户询问有效响应的功能。

简单的例子：

number = click.prompt('Please enter a number', type=float)
print(number)

Please enter a number: 
 a
Error: a is not a valid floating point value
Please enter a number: 
 10
10.0

请注意如何将字符串值自动转换为浮点数。

检查值是否在范围内：

提供了不同的custom types。要获取特定范围内的数字，我们可以使用IntRange：

age = click.prompt("What's your age?", type=click.IntRange(1, 120))
print(age)

What's your age?: 
 a
Error: a is not a valid integer
What's your age?: 
 0
Error: 0 is not in the valid range of 1 to 120.
What's your age?: 
 5
5

我们还可以只指定其中一个限制min或max：

age = click.prompt("What's your age?", type=click.IntRange(min=14))
print(age)

What's your age?: 
 0
Error: 0 is smaller than the minimum valid value 14.
What's your age?: 
 18
18

会员资格测试：

使用click.Choice类型。默认情况下，此检查区分大小写。

choices = {'apple', 'orange', 'peach'}
choice = click.prompt('Provide a fruit', type=click.Choice(choices, case_sensitive=False))
print(choice)

Provide a fruit (apple, peach, orange): 
 banana
Error: invalid choice: banana. (choose from apple, peach, orange)
Provide a fruit (apple, peach, orange): 
 OrAnGe
orange

使用路径和文件：

使用click.Path类型，我们可以检查现有路径并解决它们：

path = click.prompt('Provide path', type=click.Path(exists=True, resolve_path=True))
print(path)

Provide path: 
 nonexistent
Error: Path "nonexistent" does not exist.
Provide path: 
 existing_folder
'/path/to/existing_folder

可以通过click.File完成文件的读写：

file = click.prompt('In which file to write data?', type=click.File('w'))
with file.open():
    file.write('Hello!')
# More info about `lazy=True` at:
# https://click.palletsprojects.com/en/7.x/arguments/#file-opening-safety
file = click.prompt('Which file you wanna read?', type=click.File(lazy=True))
with file.open():
    print(file.read())

In which file to write data?: 
         # <-- provided an empty string, which is an illegal name for a file
In which file to write data?: 
 some_file.txt
Which file you wanna read?: 
 nonexistent.txt
Error: Could not open file: nonexistent.txt: No such file or directory
Which file you wanna read?: 
 some_file.txt
Hello!

其他示例：

密码确认：

password = click.prompt('Enter password', hide_input=True, confirmation_prompt=True)
print(password)

Enter password: 
 ······
Repeat for confirmation: 
 ·
Error: the two entered values do not match
Enter password: 
 ······
Repeat for confirmation: 
 ······
qwerty

默认值：

在这种情况下，只需按 Enter （或您使用的任何键）而不输入值，即可得到默认值：

number = click.prompt('Please enter a number', type=int, default=42)
print(number)

Please enter a number [42]: 
 a
Error: a is not a valid integer
Please enter a number [42]: 

42

Answer 11

Functional approach或“ 看起来没有循环！”：

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

或者如果您想将“错误输入”消息与输入提示分开，如其他答案所示：

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

它如何工作？

prompts = chain(["Enter a number: "], repeat("Not a number! Try again: "))

itertools.chain和itertools.repeat的组合将创建一个迭代器它将一次生成字符串"Enter a number: "，并无限次数生成"Not a number! Try again: "：

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)-此处map将把上一步中的所有prompts字符串应用于input函数。例如。：

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...

我们使用filter和str.isdigit过滤掉那些仅包含数字的字符串：

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...

为了只获取第一个数字字符串，我们使用next。

其他验证规则：

字符串方法：当然，您可以使用其他字符串方法，例如str.isalpha仅获得字母字符串，或str.isupper仅获得大写字母。有关完整列表，请参见docs。

成员资格测试：
有几种不同的执行方法。其中之一是使用__contains__方法：

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

数字比较：
我们可以在这里使用有用的比较方法。例如，对于__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

或者，如果您不喜欢它，则可以随时定义自己的函数，或使用operator模块中的函数。

路径存在：
这里可以使用pathlib库及其Path.exists方法：

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

尝试次数上限：

如果您不想无限次地问某人来折磨他，可以在通话itertools.repeat中指定一个限制。这可以与为next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!

合并验证规则：

例如，在一个简单的情况下，当程序要求输入1到120岁之间的年龄时，可以只添加另一个filter：

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，最好实现执行logical conjunction的函数。在下面的示例中，我将使用here中的一个现成的：

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案例提供自定义消息，那么恐怕就没有 pretty 功能方法。或者，至少我找不到。

Answer 12

使用递归功能的持久用户输入：

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)

Answer 13

好问题！您可以尝试以下代码。 =）

此代码使用ast.literal_eval() 查找输入的数据类型（age）。然后遵循以下算法：

要求用户输入他/他的age。

1.1。如果age是float或int数据类型：



检查是否为age>=18。如果为age>=18，则输出适当的输出并退出。



检查是否为0<age<18。如果为0<age<18，则输出适当的输出并退出。



如果为age<=0，请用户再次输入有效的年龄编号（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（ ie 返回步骤1。）

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.")

Answer 14

下面的代码可能有帮助。

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='')

如果要尝试最多尝试3次，请使用以下代码

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：这使用递归。

Answer 15

使用定制的ValidationError和整数输入的（可选）范围验证来使用输入验证的另一种解决方案：

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

Answer 16

使用try-except处理错误并再次重复：

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")

Answer 17

虽然try / except块可以使用，但完成此任务的更快更干净的方法是使用str.isdigit()。

while True:
    age = input("Please enter your age: ")
    if age.isdigit():
        age = int(age)
        break
    else:
        print("Invalid number '{age}'. Try again.".format(age=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.")

Answer 18

使用 isdigit() 检查字符串是否代表有效整数。

您可以使用递归函数。

def ask():
    answer = input("Please enter amount to convert: ")
    if not answer.isdigit():
        print("Invalid")
        return ask()

    return int(answer)

Gdp = ask()

或者一个while循环

while True:
    answer = input("Please enter amount to convert: ")
    if not answer.isdigit():
        print("Invalid")
        continue

    Gbp = int(answer)

Answer 19

您始终可以应用简单的if-else逻辑，并在代码中添加一个if循环以及一个for循环。

while True:
     age = int(input("Please enter your age: "))
     if (age >= 18)  : 
         print("You are able to vote in the United States!")
     if (age < 18) & (age > 0):
         print("You are not able to vote in the United States.")
     else:
         print("Wrong characters, the input must be numeric")
         continue

这将是一个无限次的厕所，您将被要求无限期地输入年龄。

Answer 20

这是一个更干净，更通用的解决方案，避免了重复的if / else块：编写一个在字典中使用（错误，错误提示）对的函数，并使用断言进行所有值检查。

def validate_input(prompt, error_map):
    while True:
        try:
            data = int(input(prompt))
            # Insert your non-exception-throwing conditionals here
            assert data > 0
            return data
        # Print whatever text you want the user to see
        # depending on how they messed up
        except tuple(error_map.keys()) as e:
            print(error_map[type(e)])

用法：

d = {ValueError: 'Integers only', AssertionError: 'Positive numbers only', 
     KeyboardInterrupt: 'You can never leave'}
user_input = validate_input("Positive number: ", d)

Answer 21

您可以将输入语句设置为True循环，以便它反复询问用户输入，然后在用户输入您想要的响应时中断该循环。并且您可以使用try和except块来处理无效响应。而真： var = True 尝试： age = int（输入（“请输入您的年龄：”））除了ValueError：打印（“输入无效。”） var = False 如果var == True：如果年龄> = 18： print（“你可以在美国投票。”）打破其他： print（“你无法在美国投票。”） var变量只是如果用户输入字符串而不是整数，程序将不会返回“您无法在美国投票”。

Answer 22

使用“while”语句直到用户输入真值，如果输入值不是数字或者是空值，则跳过它并尝试再次询问，依此类推。在示例中，我试图回答真正的问题。如果我们假设我们的年龄在1到150之间，那么接受输入值，否则它是错误的值。对于终止程序，用户可以使用0键并将其作为值输入。

注意：阅读注释顶部的代码。

{{1}}

Answer 23

您可以编写更多通用逻辑，以允许用户仅输入特定次数，因为在许多实际应用程序中出现相同的用例。

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.

询问用户输入，直到他们给出有效的响应

23 个答案:

当您的输入可能会引发异常时

实施您自己的验证规则

结合异常处理和自定义验证

将其全部封装在函数中

全部放在一起

常见的陷阱，以及为什么要避免它们

冗余使用冗余`input`语句

递归会打击你的堆栈

关注点分离

要求输入

示例

Python 3.8+注意事项

使用Click：

检查值是否在范围内：

会员资格测试：

使用路径和文件：

其他示例：

密码确认：

默认值：

Functional approach或“ 看起来没有循环！”：

它如何工作？

其他验证规则：

尝试次数上限：

合并验证规则：

字符串

整数

询问用户输入，直到他们给出有效的响应

23 个答案:

当您的输入可能会引发异常时

实施您自己的验证规则

结合异常处理和自定义验证

将其全部封装在函数中

全部放在一起

常见的陷阱，以及为什么要避免它们

冗余使用冗余input语句

递归会打击你的堆栈

关注点分离

要求输入

示例

Python 3.8+注意事项

使用Click：

检查值是否在范围内：

会员资格测试：

使用路径和文件：

其他示例：

密码确认：

默认值：

Functional approach或“ 看起来没有循环！”：

它如何工作？

其他验证规则：

尝试次数上限：

合并验证规则：

字符串

整数

冗余使用冗余`input`语句