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)?
使用自定义ValidationError和(可选的)整数输入范围验证的输入验证的另一个解决方案:
class ValidationError(ValueError):
"""Special validation error - its message is supposed to be printed"""
pass
def RangeValidator(text,num,r):
"""Generic validator - raises 'text' as ValidationError if 'num' not in range 'r'."""
if num in r:
return num
raise ValidationError(text)
def ValidCol(c):
"""Specialized column validator providing text and range."""
return RangeValidator("Columns must be in the range of 0 to 3 (inclusive)",
c, range(4))
def ValidRow(r):
"""Specialized row validator providing text and range."""
return RangeValidator("Rows must be in the range of 5 to 15(exclusive)",
r, range(5,15))
用法:
def GetInt(text, validator=None):
"""Aks user for integer input until a valid integer is given. If provided,
a 'validator' function takes the integer and either raises a
ValidationError to be printed or returns the valid number.
Non integers display a simple error message."""
print()
while True:
n = input(text)
try:
n = int(n)
return n if validator is None else validator(n)
except ValueError as ve:
# prints ValidationErrors directly - else generic message:
if isinstance(ve, ValidationError):
print(ve)
else:
print("Invalid input: ", n)
column = GetInt("Pleased enter column: ", ValidCol)
row = GetInt("Pleased enter row: ", ValidRow)
print( row, column)
输出:
Pleased enter column: 22
Columns must be in the range of 0 to 3 (inclusive)
Pleased enter column: -2
Columns must be in the range of 0 to 3 (inclusive)
Pleased enter column: 2
Pleased enter row: a
Invalid input: a
Pleased enter row: 72
Rows must be in the range of 5 to 15(exclusive)
Pleased enter row: 9
9, 2
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.")
下面的代码可能会有所帮助。
age=(lambda i,f: f(i,f))(input("Please enter your age: "),lambda i,f: i if i.isdigit() else f(input("Please enter your age: "),f))
print("You are able to vote in the united states" if int(age)>=18 else "You are not able to vote in the united states",end='')
如果你想要最大尝试次数,比如3次,请使用下面的代码
age=(lambda i,n,f: f(i,n,f))(input("Please enter your age: "),1,lambda i,n,f: i if i.isdigit() else (None if n==3 else f(input("Please enter your age: "),n+1,f)))
print("You are able to vote in the united states" if age and int(age)>=18 else "You are not able to vote in the united states",end='')
注意:这里使用递归。