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

Answer 1

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

Answer 2

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.

Answer 3

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)