[Tutor] A simple "if" and "elif" problem
xgermx at gmail.com
Sat May 5 20:36:51 CEST 2012
Python novice here. Caveat emptor.
Since all python input is read in as a string, you could try checking
to see if the value is a digit with the .isdigit string method.
On Sat, May 5, 2012 at 1:08 PM, Emile van Sebille <emile at fenx.com> wrote:
> On 5/5/2012 11:01 AM Santosh Kumar said...
>> I am reading the documentation and I'm in the section 4.1. Let me
>> write it down here:
> You'll need to peek ahead to section 8 Errors and Exceptions.
> Try and see if that doesn't get you going.
>>>>> x = int(input("Please enter an integer: "))
>> Please enter an integer: 42
>>>>> if x< 0:
>> ... x = 0
>> ... print('Negative changed to zero')
>> ... elif x == 0:
>> ... print('Zero')
>> ... elif x == 1:
>> ... print('Single')
>> ... else:
>> ... print('More')
>> Now I want to add a filter in this script, I want when a user enter a
>> string here it give a warning "Please enter a number like 0 or 2".
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