CODEWITHSUNDEEP
pythontry-catchexcept
vishal
vishal

Reputation: 47

try except statement error python

def validnumber():
    notValid=True
    while(notValid==True):
        number=input('Enter number between 0 and 9->')
        if number=='':
            print('Empty input!')
        else:
            try:
                number=int(number)

            except ValueError:
                print('Number not an int value!Try Again!')
            else:
                if number>=0 and number<=9:
                    notvalid=False
    return number

def main():
    myvalidnumber=validnumber()
    print(myvalidnumber)

main()

Hey guys. I wrote this program and just had 1 question.

-> the program does not end even if i enter a number between 0 and 9. Could anyone explain why is this happening?.

Thanks in advance :)

Upvotes: 1

Views: 114

Answers (2)

bruno desthuilliers
bruno desthuilliers

Reputation: 77902

You already got the solution from thefourtheye. Just for the record: the canonical way to implement such a function in Python is to use an infinite loop and either a break statement or an early return (as in the example below). Also proper use of the continue statement can simplify the flow:

def validnumber():
    while True:
        number=input('Enter number between 0 and 9->')
        if number=='':
            print('Empty input!')
            continue

        try:
            number=int(number)
        except ValueError:
            print('Number not an int value! Try Again!')
            continue

        if number < 0 or number > 9:
           print('Number not between 0 and 9! Try Again!')
           continue

        return number

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239473

Python's variables are case sensitive. notvalid is not the same as notValid. So, when you say

notvalid=False

you are creating a new variable. Just change it to

notValid = False

and you are fine.

Upvotes: 2

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