How to get the name of an exception that was caught in Python?

Each Answer to this Q is separated by one/two green lines.

How can I get the name of an exception that was raised in Python?

e.g.,

try:
    foo = bar
except Exception as exception:
    name_of_exception = ???
    assert name_of_exception == 'NameError'
    print "Failed with exception [%s]" % name_of_exception

For example, I am catching multiple (or all) exceptions, and want to print the name of the exception in an error message.

Here are a few different ways to get the name of the class of the exception:

  1. type(exception).__name__
  2. exception.__class__.__name__
  3. exception.__class__.__qualname__

e.g.,

try:
    foo = bar
except Exception as exception:
    assert type(exception).__name__ == 'NameError'
    assert exception.__class__.__name__ == 'NameError'
    assert exception.__class__.__qualname__ == 'NameError'

If you want the fully qualified class name (e.g. sqlalchemy.exc.IntegrityError instead of just IntegrityError), you can use the function below, which I took from MB’s awesome answer to another question (I just renamed some variables to suit my tastes):

def get_full_class_name(obj):
    module = obj.__class__.__module__
    if module is None or module == str.__class__.__module__:
        return obj.__class__.__name__
    return module + '.' + obj.__class__.__name__

Example:

try:
    # <do something with sqlalchemy that angers the database>
except sqlalchemy.exc.SQLAlchemyError as e:
    print(get_full_class_name(e))

# sqlalchemy.exc.IntegrityError

You can also use sys.exc_info(). exc_info() returns 3 values: type, value, traceback. On documentation: https://docs.python.org/3/library/sys.html#sys.exc_info

import sys

try:
    foo = bar
except Exception:
    exc_type, value, traceback = sys.exc_info()
    assert exc_type.__name__ == 'NameError'
    print "Failed with exception [%s]" % exc_type.__name__

This works, but it seems like there must be an easier, more direct way?

try:
    foo = bar
except Exception as exception:
    assert repr(exception) == '''NameError("name 'bar' is not defined",)'''
    name = repr(exception).split('(')[0]
    assert name == 'NameError'

You can print the exception using some formated strings:

Example:

try:
    #Code to execute
except Exception as err:
    print(f"{type(err).__name__} was raised: {err}")

The other answers here are great for exploration purposes, but if the primary goal is to log the exception (including the name of the exception), perhaps consider using logging.exception instead of print?


The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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