相关疑难解决方法(0)

询问用户输入,直到他们给出有效的响应

我正在编写一个必须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.")
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如果用户输入合理数据,这将按预期工作.

C:\Python\Projects> canyouvote.py
Please enter your age: 23
You are able to vote in the United States!
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但如果他们犯了错误,那就崩溃了:

C:\Python\Projects> canyouvote.py
Please enter your age: dickety six
Traceback (most recent call last):
  File "canyouvote.py", line 1, in …
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python validation loops user-input python-3.x

523
推荐指数
10
解决办法
42万
查看次数

如何将输入读作数字?

在下面的代码中为什么xy字符串而不是整数?网上的所有内容都说要使用raw_input(),但是我读input()raw_input()在Python 3.x中重命名的Stack Overflow(在一个不处理整数输入的线程上).

play = True

while play:

    x = input("Enter a number: ")
    y = input("Enter a number: ")

    print(x + y)
    print(x - y)
    print(x * y)
    print(x / y)
    print(x % y)

    if input("Play again? ") == "no":
        play = False
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python int input python-2.7 python-3.x

261
推荐指数
7
解决办法
76万
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标签 统计

python ×2

python-3.x ×2

input ×1

int ×1

loops ×1

python-2.7 ×1

user-input ×1

validation ×1