For the simple snippet below
a = raw_input("Enter a number: ")
How can I write an exception such that if a string is entered, the user is re-prompted to try again and again until an integer is entered.
In other words, keep doing this until an integer is entered.:
if a.isdigit == False:
raw_input("Try again: ")
The usual way to achieve this is to use int() with exception handling:
>>> n = None
>>> while n is None:
... a = raw_input("Enter a number: ")
... try:
... n = int(a)
... except ValueError:
... print "Not a number."
...
Enter a number: abc
Not a number.
Enter a number: cauliflowers are my favourite vegetable
Not a number.
Enter a number: 12
>>>
However if you would prefer to avoid exception handling you can take the following approach. Note that it does not accept negative integers, but only the natural numbers 0, 1, 2, ..., N:
>>> while True:
... a = raw_input("Enter a number: ")
... if a.isdigit():
... break
... print "Not a number."
...
Enter a number: I like beetroot too
Not a number.
Enter a number: -500
Not a number.
Enter a number: 500
>>>
It becomes a little more involved to handle negative integers in this manner, requiring three cases, one for empty strings, one for negative integers and one for natural numbers:
>>> while True:
... a = raw_input("Enter a number: ")
... if len(a) > 0 and ((a[0] == '-' and a[1:].isdigit()) or a.isdigit()):
... break
... print "Not a number."
...
Enter a number: Celery stinks.
Not a number.
Enter a number:
Not a number.
Enter a number: -
Not a number.
Enter a number: 1
>>>
try...except? If an addition int(a) > 0 condition was added to the answers you downvoted, wont it make their solution valid as well?
try... except is required because int(a) returns an integer, not a boolean, and uses exceptions to signal the case where it is dealing with a string that is not a valid integer. This is why adding int(a) > 0 to the isdigit answers is not enough, as int will throw an exception if confronted with a non-integer representing string.0, 1, 2, ..., N) rather than integers (-N, ..., -2, -1, 0, 1, 2, ..., N) I would recommend updating your question and using Jeff or Michaels answers below.You can use a while loop:
a = ""
while not a.isdigit():
a = raw_input("Enter a number: ")
If you want an error message it becomes something like:
a = raw_input("Enter a number: ")
while not a.isdigit():
a = raw_input("Try again: ")
.isdigit right in his question...while True:
a = raw_input("Enter a number:")
try:
i = int(a)
break
except ValueError:
print "Not a number"
a = raw_input("Enter a number: ")
while (not a.isdigit()):
a = raw_input("Try again: ")
isdigit(), hope i did help this time :)