How to get Exception arguments? [duplicate]

Question

I was trying to find a way to do this in python 3.6.5 which is not supported

try:
     c=1/0
     print (c)
except ZeroDivisionError, args:
     print('error dividing by zero', args)

It says this type of syntax is not supported by python 3.6.5 So is there a way to get the arguments of the exception?

Answer 1

How about:

try:
     c=1/0
     print (c)
except ZeroDivisionError as e:
     print('error dividing by zero: ' + str(e.args))

Comma notation is now used to except multiple types of exceptions, and they need to be in parentheses, like:

try:
    c = int("hello")
    c = 1 / 0
    print(c)
except (ZeroDivisionError, ValueError) as e:
    print('error: ' + str(e.args))