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,程序就能正常工作。

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:

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,我希望程序再次要求输入。是这样的:

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!

我如何要求有效输入而不是崩溃或接受无效值(例如-1)?


当前回答

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."

其他回答

使用点击:

Click是一个命令行界面库,它提供了向用户请求有效响应的功能。

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

注意它是如何自动将字符串值转换为浮点数的。

检查一个值是否在一个范围内:

提供了不同的自定义类型。为了得到一个特定范围内的数字,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

我们也可以只指定一个极限,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

会员测试:

使用点击。选择类型。默认情况下,该检查是区分大小写的。

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:

使用点击。路径类型,我们可以检查现有的路径并解析它们:

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

文件的读写可以通过点击完成。文件:

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!

其他的例子:

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(或任何你使用的键)而不输入值,就会给你一个默认值:

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

所以,我最近在搞一些类似的事情,我想出了下面的解决方案,它使用一种方法来获取输入,拒绝垃圾,甚至在它以任何逻辑方式检查之前。

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()

您可以在这里找到完整的模块。

例子:

$ ./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!
$ _

注意,这个实现的本质是,一旦读取非数字的内容,它就会关闭stdin。我没有在a后面按回车键,但我需要在数字后面按。

您可以将此与同一模块中的thismany()函数合并,以只允许(比如说)三位数字。

使用isdigit()检查字符串是否代表有效的整数。

你可以使用递归函数。

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

    return int(answer)

Gdp = ask()

或者while循环

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

    Gbp = int(answer)

为什么你要做一个while True,然后跳出这个循环,而你也可以把你的要求放在while语句中因为你想要的是一旦你有了年龄就停止?

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.

这是可行的,因为年龄永远不会有一个没有意义的值,代码遵循“业务流程”的逻辑。

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."