Getting name of the exception along with the error

Question

How to get the name of the exception in try..except block of Python?

try:
    #code that throws errors
except Exception as e:
    #A broad except block to catch all errors
    #Handling the error

Sometimes when there are a lot of exceptions to handle, catching Exception seems easy (though discouraged). But I'm not able to find the name of the exception, is there any way to do it?

`print(e)` will print the string representation of the exception, `print(type(e))` will print the class of the exception — Iain Shelvington, Jan 11 '20 at 16:47

Wasif · Answer 1 · 2020-01-11T16:52:08.137

2

Use this:

type(e).__name__

Or

type(e).__class__.name

Or

type(e).__class__.qualname

edited Jan 11 '20 at 16:52

answered Jan 11 '20 at 16:46

Wasif

14,755
3
14
34

Exceptions do not have an attribute `__name__` - `type(e).__name__` would work however – Iain Shelvington Jan 11 '20 at 16:51
All the three didn't work. `AttributeError`. :) But hey, `e.__name__` works (its same as `type(e)` I suppose?) – J Arun Mani Jan 11 '20 at 16:52
1

No not quite same..... – Wasif Jan 11 '20 at 16:52
Oops sorry. Thought `type` returns `__name__` of an object. – J Arun Mani Jan 11 '20 at 16:58

score 1 · Accepted Answer · answered Jan 11 '20 at 16:41

1

Assuming by "name" you mean "type", try type(e) in your except block.

answered Jan 11 '20 at 16:41

ChrisGPT was on strike

127,765
105
273
257

Getting name of the exception along with the error

2 Answers2