raise custom exception in python

Question

I could not get what is wrong with the code ? when I execute nothing happens. I am expecting my custom error message.

def testing():
  try:
    raise Exception('My error!')
  except:
    pass

testing()

I wanted to raise a custom exception. so I did it in try block and left except block to do nothing — Hamsavardhini
– Hamsavardhini, Commented Oct 1, 2020 at 15:35

F.NiX · Accepted Answer · 2020-09-30 19:37:10Z

5

You are raising an exception successfully. But you are catching that with try/except block. So nothing happens unless you describe it in except block.

answered Sep 30, 2020 at 19:37

F.NiX

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Comments

Goldwave · Accepted Answer · 2020-09-30 19:37:25Z

4

You are successfully raising an error. And the try/catch statements sees it, and goes to catch as you have raised an error.

To fully customize errors you can declare them as so:

class CustomError(Exception):
    pass
raise CustomError("An error occurred")

results in

__main__.CustomError: An error occurred

answered Sep 30, 2020 at 19:37

Goldwave

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1 Comment

Mark Ransom Over a year ago

You can also define an __init__ function so that there's no need to pass an error string, the class can provide it.

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