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

您总是可以应用简单的if-else逻辑,并在for循环的同时向代码中添加一个if逻辑。

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

这将是一个无限的厕所,你将被要求进入一个无限的时代。

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

您可以将input语句设置为while True循环,这样它就会重复请求用户输入,然后在用户输入您想要的响应时打破循环。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变量是这样的,如果用户输入一个字符串而不是一个整数,程序不会返回“你不能在美国投票”。

我是Unix哲学“只做一件事并把它做好”的忠实粉丝。捕获用户输入并验证它是两个独立的步骤:

使用get_input提示用户输入,直到输入成功 使用可以传递给get_input的验证器函数进行验证

它可以保持简单如(Python 3.8+,使用walrus操作符):

def get_input(
    prompt="Enter a value: ",
    validator=lambda x: True,
    error_message="Invalid input. Please try again.",
):
    while not validator(value := input(prompt)):
        print(error_message)
    return value

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

if __name__ == "__main__":
    val = get_input("Give a positive number: ", is_positive_int)
    print(f"OK, thanks for {val}")

示例运行:

Give a positive number: -5
Invalid input. Please try again.
Give a positive number: asdf
Invalid input. Please try again.
Give a positive number:
Invalid input. Please try again.
Give a positive number: 42
OK, thanks for 42

在Python < 3.8中,你可以像这样使用get_input:

def get_input(
    prompt="Enter a value: ",
    validator=lambda x: True,
    error_message="Invalid input. Please try again.",
):
    while True:
        value = input(prompt)
        if validator(value):
            return value
        print(error_message)

您还可以在终止应用程序之前处理KeyboardInterrupt并打印友好的退出消息。如果需要,可以使用计数器限制允许的重试次数。

虽然try/except块可以工作,但是使用str.isdigit()可以更快更清晰地完成此任务。

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try-except来处理错误并重复一次:

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")