How to print the stack trace of an exception object in Python?
Note that the question is NOT about printing stack trace of LAST exception. Exception object may be saved at some distant point of time in the past.
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2Similar: stackoverflow.com/questions/3702675/…user202729– user2027292018-10-10 14:48:22 +00:00Commented Oct 10, 2018 at 14:48
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2 Answers
It's a bit inconvenient, but you can use traceback.print_exception. Given an exception ex:
traceback.print_exception(type(ex), ex, ex.__traceback__)
Example:
import traceback
try:
1/0
except Exception as ex:
traceback.print_exception(type(ex), ex, ex.__traceback__)
# output:
# Traceback (most recent call last):
# File "untitled.py", line 4, in <module>
# 1/0
# ZeroDivisionError: division by zero
6 Comments
Jean-François Fabre
It's a bit inconvenient why? looks okay to me
Aran-Fey
@Jean-FrançoisFabre Well, there really is no reason why it's necessary to pass 3 arguments to that function. If a function like
print_exception(ex) existed, that would be much nicer.
Alex Pshenko
Other useful function would be
print_exc() which get info about exception(s) from sys.info
matanox
See also
traceback.format_exc()
Nishant
IMHO,
exception.traceback() is a more cleaner OOPs-like syntax. |
you can manually iterate through the __traceback__ attribute to print lines & files:
def function():
raise ValueError("flfkl")
try:
function()
except Exception as e:
traceback = e.__traceback__
while traceback:
print("{}: {}".format(traceback.tb_frame.f_code.co_filename,traceback.tb_lineno))
traceback = traceback.tb_next
answered Oct 10, 2018 at 14:46