python - 要求用户输入，直到他们给出有效的响应

Question

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，程序就会按预期运行。

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就会失败:

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:

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!

如何请求有效输入而不是崩溃或接受无效值(例如 -1)？

Answer 1

完成此操作的最简单方法是将 input 方法置于 while 循环中。使用continue当您输入错误时，当您满意时 break 退出循环。

当您的输入可能引发异常时

使用 try and 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.")

将其全部封装在一个函数中

如果您需要向用户询问很多不同的值，将这段代码放在一个函数中可能会很有用，这样您就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: ")

把它们放在一起

你可以扩展这个想法来制作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

有如下用法:

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input语句的冗余使用

此方法有效，但通常被认为风格不佳:

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): ")

它最初可能看起来很吸引人，因为它比 while True 方法短，但它违反了 Don't Repeat Yourself软件开发原理。这增加了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，但不小心只更改了上面的第一个 input 怎么办？这是一个等待发生的SyntaxError。

递归会破坏你的堆栈

如果您刚刚了解了递归，您可能会想在 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能正常工作，但如果用户输入无效数据的次数足够多，脚本将终止并出现 RuntimeError: maximum recursion depth exceeded。你可能认为“没有傻子会连续犯 1000 个错误”，但你低估了傻子的聪明才智!