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 = ???
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__
Here are a few different ways to get the name of the class of the exception:
type(exception).__name__
exception.__class__.__name__
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
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'
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?