请求用户进行输入，直到他们给出有效的响应

Question

我正在编写一个程序，接受用户的输入。

#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，程序就会按预期工作。

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

但如果用户输入无效数据，则失败：

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'

我希望程序再次请求输入，而不是崩溃。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!

当输入非感官数据时，如何使程序请求有效的输入而不是崩溃？

我怎样才能拒绝像-1这样的值，这是一个有效的int，但在这个上下文中是没有意义的？

Answer 1

最简单的方法是将input方法放入while循环中。当输入错误时使用^{}，当您满意时使用break退出循环。

当您的输入可能引发异常时

使用^{} and ^{}检测用户何时输入无法分析的数据。

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

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.")

将其全部封装在函数中

如果需要向用户询问许多不同的值，那么将此代码放入函数中可能会很有用，因此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: ")

总而言之

您可以将此思想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'))

常见的陷阱，以及为什么要避免它们

冗余语句的冗余使用

这种方法有效，但通常被认为是低劣的风格：

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，但它违反了软件开发的Don't Repeat Yourself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，但意外地只更改上面的第一个input，将backport更改为2.7，该怎么办？只是在等待发生。

递归将破坏堆栈

如果您刚刚学习了递归，那么您可能会尝试在get_non_negative_int中使用它，以便可以处理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，但是如果用户输入无效数据的次数足够多，脚本将以RuntimeError: maximum recursion depth exceeded结束。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！

Answer 2

尽管公认的答案是惊人的。我还想分享一个快速破解这个问题的方法。（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: ")))

这段代码是针对python 3.x的

Answer 3

为什么你要做一个while True，然后跳出这个循环，而你也可以把你的需求放在while语句中，因为你只想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一个没有意义的价值，代码遵循您的“业务流程”逻辑