How do I log a Python error with debug information?


Question

I am printing Python exception messages to a log file with logging.error:

import logging
try:
    1/0
except ZeroDivisionError as e:
    logging.error(e)  # ERROR:root:division by zero

Is it possible to print more detailed information about the exception and the code that generated it than just the exception string? Things like line numbers or stack traces would be great.

1
393
5/23/2018 3:16:28 PM

Accepted Answer

logger.exception will output a stack trace alongside the error message.

For example:

import logging
try:
    1/0
except ZeroDivisionError as e:
    logging.exception("message")

Output:

ERROR:root:message
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
ZeroDivisionError: integer division or modulo by zero

@Paulo Cheque notes, "be aware that in Python 3 you must call the logging.exception method just inside the except part. If you call this method in an arbitrary place you may get a bizarre exception. The docs alert about that."

623
5/23/2018 3:15:43 PM

One nice thing about logging.exception that SiggyF's answer doesn't show is that you can pass in an arbitrary message, and logging will still show the full traceback with all the exception details:

import logging
try:
    1/0
except ZeroDivisionError:
    logging.exception("Deliberate divide by zero traceback")

With the default (in recent versions) logging behaviour of just printing errors to sys.stderr, it looks like this:

>>> import logging
>>> try:
...     1/0
... except ZeroDivisionError:
...     logging.exception("Deliberate divide by zero traceback")
... 
ERROR:root:Deliberate divide by zero traceback
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
ZeroDivisionError: integer division or modulo by zero

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