[passer_by]

Hi guys,

I want to do an input check to the characters that the user enters .
The only permitted inputs are Integers , i.e. ,if the user enters a series
that has at least one Space (" ") or at least one character other than the digits 0-9
then the Input above is incorrect.

So why not do it with

choice=input()

?
In that way I get a string (str) , right ? but how can i check each and every character
in the string ? because the string is not a list ,and isn't an array .

ideas ?
Thanks,Ron

[Nallo]

passer_by, on 11 April 2011 - 02:44 AM, said:

but how can i check each and every character in the string ?

You can loop through a string just like you loop through a list:

>>> for c in "kli":
	print(c)

k
l
i

and ckeck those characters to see if they are in "0123456789"

>>> "a" in "01234567890"
False
>>> "4" in "0123456789"
True

[Motoma]

Another option would be to cast to an integer and catch a ValueError exception.

choice = input()
try:
  choice = int(choice)
except ValueError as ex:
  print("Bad input")

[JackOfAllTrades]

How about using isdigit on the entered input?

choice = 'a'
while not choice.isdigit():
    choice = input()

[baavgai]

I prefer Motoma's option. I'd put it in a loop with a "try again" kind of thing:

def getNum():
	while True:
		try:
			return int(input())
		except ValueError:
			print("Sorry, not number data, please try again")

choice = getNum()

This post has been edited by baavgai: 11 April 2011 - 06:52 AM

[JackOfAllTrades]

I was thinking of avoiding the overhead of exception handling. Probably not a big deal in this case.

[passer_by]

Hi guys ,

Thank you all first of all . Second ,can you please explain the difference between :

choice=""
and 
choice=''

?

thanks again
Ron

[Dogstopper]

No functional difference:
http://docs.python.o...string-literals