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)?
使用“while”语句，直到用户输入一个真值，如果输入值不是一个数字或它是一个空值跳过它，并尝试再次询问，等等。
举例来说，我试图真正地回答你的问题。如果我们假设我们的年龄在1到150之间，那么输入值被接受，否则它是一个错误的值。
对于终止程序，用户可以使用0键并输入它作为一个值。
注意:阅读代码顶部的注释。
# If your input value is only a number then use "Value.isdigit() == False".
# If you need an input that is a text, you should remove "Value.isdigit() == False".
def Input(Message):
Value = None
while Value == None or Value.isdigit() == False:
try:
Value = str(input(Message)).strip()
except Exception:
Value = None
return Value
# Example:
age = 0
# If we suppose that our age is between 1 and 150 then input value accepted,
# else it's a wrong value.
while age <=0 or age >150:
age = int(Input("Please enter your age: "))
# For terminating program, the user can use 0 key and enter it as an a value.
if age == 0:
print("Terminating ...")
exit(0)
if age >= 18 and age <=150:
print("You are able to vote in the United States!")
else:
print("You are not able to vote in the United States.")
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。
好问题!您可以尝试以下代码。=)
这段代码使用ast.literal_eval()来查找输入的数据类型(age)。然后按照以下算法:
请用户输入年龄。
1.1. 如果age为float或int数据类型:
检查年龄>是否=18。如果age>=18，打印相应的输出并退出。
检查0<年龄<18。如果0<age<18，打印适当的输出并退出。
如果age<=0，请用户再次输入age的有效数字(即返回步骤1)。
1.2. 如果age不是float或int数据类型，则要求用户再次输入她/他的年龄(即返回第1步)。
这是代码。
from ast import literal_eval
''' This function is used to identify the data type of input data.'''
def input_type(input_data):
try:
return type(literal_eval(input_data))
except (ValueError, SyntaxError):
return str
flag = True
while(flag):
age = raw_input("Please enter your age: ")
if input_type(age)==float or input_type(age)==int:
if eval(age)>=18:
print("You are able to vote in the United States!")
flag = False
elif eval(age)>0 and eval(age)<18:
print("You are not able to vote in the United States.")
flag = False
else: print("Please enter a valid number as your age.")
else: print("Sorry, I didn't understand that.")
