我正在编写一个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.")
只要用户输入有意义的数据，该程序就会按预期运行。
C:\Python\Projects> canyouvote.py
Please enter your age: 23
You are able to vote in the United States!
但是如果用户输入无效数据，它将失败：
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'
除了崩溃，我希望程序再次请求输入。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!
我如何使程序要求输入有效信息，而不是在输入无意义的数据时崩溃？
我该如何拒绝像-1这样的值，该值是有效的int，但在这种情况下是荒谬的？
最简单的方法是将input方法放入while循环中。使用 输入错误时继续 ，并在满意时中断。
使用 try和except 来检测用户何时输入了无法解析的数据。
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.")
如果要拒绝Python可以成功解析的值，则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
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.")
如果您需要询问用户许多不同的值，则可以将此代码放入函数中，因此不必每次都重新输入。
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: ")
您可以将这个想法扩展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
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'))
input语句此方法有效，但通常被认为是较差的样式：
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): ")
最初看起来很有吸引力，因为它比whileTrue方法短，但是它违反了不要重复自己的软件开发原理。这增加了系统中错误的可能性。如果您想通过将input更改为raw_input来反向移植到2.7，而又意外地仅更改了上面的第一个input，该怎么办？这是一个SyntaxError，正等着发生。
如果您刚刚了解了递归，则可能会想在get_non_negative_int中使用它，以便处理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可以正常工作，但是如果用户输入无效数据的次数足够多，该脚本将以RuntimeError:超出最大递归深度终止。您可能会认为"没有傻瓜会连续犯1000个错误"，但是您却低估了傻瓜的聪明才智！
为什么要先做一个whileTrue，然后跳出这个循环，而您也可以只将需求放在while语句中，因为您想要的只是一旦年龄就停止了？ / p>
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.
这将起作用，因为年龄永远不会有没有意义的值，并且代码遵循您的"业务流程"的逻辑
尽管公认的答案是惊人的。我也想分享一个快速解决此问题的方法。 (这也解决了负面的年龄问题。)
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。这段代码适用于python 3.x。
因此，我最近在搞些类似的事情，于是我想到了以下解决方案，该解决方案使用了一种获取垃圾输入的方式，甚至可以以任何逻辑方式对其进行检查。
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()
您可以在此处找到完整的模块。
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，一旦读取到不是数字的内容，它将关闭stdin。我没有在a之后按回车键，但是我需要在数字后面输入。
您可以将其与thismany()函数合并到同一模块中，以仅允许输入三位数。
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
prompts = chain(["Enter a number: "], repeat("Not a number! Try again: "))
itertools.chain的组合 和 itertools.repeat 将创建一个迭代器，该迭代器将生成字符串"输入数字:"一次，而"不是数字!请重试:"无限次： 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(输入,提示)-此处 map 会将上一步中的所有提示字符串应用于 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...
filter 和 str.isdigit 来过滤出仅包含那些字符串数字：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...
下一个 。 字符串方法：当然，您可以使用其他字符串方法，例如 str.isalpha 仅获取字母字符串，或 str.isupper 仅获取大写字母。有关完整列表，请参见文档。
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：
这里有一些有用的比较方法。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，如果您不喜欢使用dunder方法(dunder =双下划线)，则可以始终定义自己的函数，或使用 operator 模块。
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
如果您不想无限次折磨用户，可以在 itertools.repeat 。这可以与为 next 提供默认值结合使用a>函数：
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!
有时候，如果用户不小心输入了 IN CAPS 或在字符串的开头或结尾都带有空格，我们就不想拒绝输入。要考虑这些简单的错误，我们可以通过应用 str来预处理输入数据.lower 和 str.strip 方法。例如，对于成员资格测试，代码如下所示：
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
如果要使用许多函数进行预处理，则使用执行功能组成。例如，使用此处中的一个：
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
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，只需添加另一个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，最好实现执行逻辑合取。在下面的示例中，我将使用此处：
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，如果有人为每个失败的案例需要自定义消息，那么恐怕就没有 pretty 功能。或者，至少我找不到一个。
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：
import click
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。
提供了不同的自定义类型。要获得特定范围内的数字，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
我们还可以仅指定一个限制，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
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：
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!
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
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
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."
在Daniel Q和Patrick Artner的出色建议的基础上，这是一个更为通用的解决方案。
# 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)
我选择了显式的if和raise语句，而不是assert，因为可以关闭断言检查，而应该始终进行验证可以提供强大的功能。
这可以用于获取具有不同验证条件的不同种类的输入。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.")
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try / except块将起作用，但是完成此任务的更快，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.")
使用自定义的ValidationError和用于整数输入的(可选)范围验证的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
好问题！您可以尝试以下代码。 =)
此代码使用 ast.literal_eval()来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
年龄。1.1。如果
age是float或int数据类型：
检查是否
age>=18。如果age>=18，则打印适当的输出并退出。检查是否
0。如果<18 0，则打印适当的输出并退出。<18 如果
age<=0，请要求用户再次输入有效的年龄编号，( ie 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其年龄( 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.")
使用递归功能的持久用户输入：
def askName():
return input("Write your name: ").strip() or askName()
name = askName()
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)
您始终可以应用简单的if-else逻辑，并在代码中添加一个if逻辑以及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。
您可以编写更通用的逻辑，以允许用户仅输入特定的次数，因为在许多实际应用程序中会出现相同的用例。
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.
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。而且，您可以使用try和except块来处理无效响应。
while True:
var = True
try:
age = int(input("Please enter your age: "))
except ValueError:
print("Invalid input.")
var = False
if var == True:
if age >= 18:
print("You are able to vote in the United States.")
break
else:
print("You are not able to vote in the United States.")
var变量就是这样，如果用户输入一个字符串而不是一个整数，程序将不会返回"您无法在美国投票"。
使用" while"语句，直到用户输入一个真值，并且如果输入值不是数字或它是空值，请跳过该语句并尝试再次询问，依此类推。例如，我试图真正回答您的问题。如果我们认为年龄在1到150之间，则接受输入值，否则输入的值是错误的。对于终止程序，用户可以使用0键并将其作为值输入。
注意：请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 InputError:
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.")
这是一个更简洁，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)
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
上面的代码说明：为了使年龄有效，该年龄应该为正，并且不得超过正常的身体年龄，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，我们认为该输入无效，然后要求用户重试。
输入有效输入后，我们将执行检查(使用嵌套的if-else语句)，确定年龄是否大于等于18，反之亦然，并显示一条消息，说明用户是否有资格投票