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
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为什么'=='在Python之前'in'出现?

以下代码输出False,根据Python操作顺序,它应该输出True(顺序应该在 - > ==,而不是相反).为什么==之前会来?

y = "33"
"3" in y == True
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产量

False
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python

7
推荐指数
2
解决办法
232
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标签 统计

python ×2

loops ×1

python-3.x ×1

user-input ×1

validation ×1