ValueError Exception handling in Python

Question

I want to handle ValueError in Python. But whenever I type the below code I get an error and I want the output as I excepted.

a, b = map(int, input().split())
try:
    print(a//b)
except ZeroDivisionError as e:
    print('Enter code: ', e)
except ValueError as e:
    print('Enter code: ', e)

If my inputs for a and b are '1' and '$' than Expected output for ValueError: 'Enter code: invalid literal for int() with base 10: $

Does this answer your question? [ValueError: invalid literal for int() with base 10: ''](https://stackoverflow.com/questions/1841565/valueerror-invalid-literal-for-int-with-base-10) — AMC, Mar 20 '20 at 18:52

score 2 · Answer 1 · answered Mar 20 '20 at 18:40

2

Your problem is before your try, here a, b = map(int, input().split())

The $ is given to int, this fails and raise the invalid literal for int() with base 10: $ which is pretty explicit

try:
    a, b = map(int, input().split())
    print(a//b)
except ZeroDivisionError as e:
    print('Enter code: ', e)
except ValueError as e:
    print('Enter code: ', e)

answered Mar 20 '20 at 18:40

azro

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Thanks, it really helped me – Sharma Shivlal Mar 20 '20 at 18:45
@ShivManoharlalSharma Think about reading the Exception, they are helpful ;) – azro Mar 20 '20 at 18:47
@ShivManoharlalSharma You can now think about [accepting the answer](https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work) – azro Mar 21 '20 at 08:52

ValueError Exception handling in Python

1 Answers1