Can I make Python output exceptions in one line / via logging?

Question

I am using AWS and use AWS cloudwatch to view logs. While things should not break on AWS, they could. I just had such a case. Then I searched for Traceback and just got the lines

Traceback (most recent call last):

without the actual traceback. I have a working structured logging setup (see other question) and I would like to get tracebacks in a similar way.

So instead of:

Traceback (most recent call last):
  File "/home/math/Desktop/test.py", line 32, in <module>
    adf
NameError: name 'adf' is not defined

something like

{"message": "Traceback (most recent call last):\n      File \"/home/math/Desktop/test.py\", line 32, in <module>\n        adf\n    NameError: name 'adf' is not defined", "lineno": 35, "pathname": "/home/math/Desktop/test.py"}

or even better also with the string in a JSON format.

The only way to achieve this I can think of is a giant try-except block. Pokemon-style. Is there a better solution?

So you don't want a try except block and also want to log errors with formatted way? — Arpit Solanki
– Arpit Solanki, Commented Feb 6, 2018 at 11:40
You should use a custom logging.Formatter and implement formatException. — Giacomo Alzetta
– Giacomo Alzetta, Commented Feb 6, 2018 at 11:42

Martin Thoma · Accepted Answer · 2018-02-06 13:08:06Z

21

You can use sys.excepthook. It is invoked whenever an exception occurs in your script.

import logging
import sys
import traceback

def exception_logging(exctype, value, tb):
    """
    Log exception by using the root logger.

    Parameters
    ----------
    exctype : type
    value : NameError
    tb : traceback
    """
    write_val = {'exception_type': str(exctype),
                 'message': str(traceback.format_tb(tb, 10))}
    logging.exception(str(write_val))

Then in your script you have to override the value of sys.excepthook.

sys.excepthook = exception_logging

Now whenever an exception occurs it will be logged with your logger handler.

Note: Don't forget to setup logger before running this

edited Feb 6, 2018 at 13:08

Martin Thoma

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answered Feb 6, 2018 at 12:55

Arpit Solanki

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1 Comment

benzkji Over a year ago

I'm trying this with my django application...any idea where to place it, codewise? I'm trying a apps.py and AppConfig, using ready() to override the sys.excepthook. As seen in the following link, logging is setup quite early, so, before ready....but no look, still many lines...docs.djangoproject.com/en/3.0/ref/applications/…

Attila123 · Accepted Answer · 2018-10-05 17:54:49Z

4

In case somebody wants the exception logged in its default format, but in one line (for any reason), based on the accepted answer:

def exception_logging(exctype, value, tb):
    """
    Log exception in one line by using the root logger.

    Parameters
    ----------
    exctype : exception type
    value : seems to be the Exception object (with its message)
    tb : traceback
    """
    logging.error(''.join(traceback.format_exception(exctype, value, tb)))

Please also note, that it uses logging.error() instead of logging.exception() which also printed some extra "NoneType: None" line.
Also note that it only seems to work with uncaught exceptions.
For logging caught exceptions, visit How do I can format exception stacktraces in Python logging? and see also my answer.

edited Oct 5, 2018 at 17:54

answered Oct 5, 2018 at 16:11

Attila123

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Comments

Martin Thoma · Accepted Answer · 2019-05-20 12:14:20Z

0

A slight variation: If you run a Flask application, you can do this:

@app.errorhandler(Exception)
def exception_logger(error):
    """Log the exception."""
    logger.exception(str(error))
    return str(error)

answered May 20, 2019 at 12:14

Martin Thoma

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