How do I log a Python error with debug information

0 votes

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.

May 11, 2020 in Python by kartik
• 37,520 points
1,204 views

You could probably use traceback for this.

import traceback

def log_traceback(ex, ex_traceback=None):
    if ex_traceback is None:
        ex_traceback = ex.__traceback__
    tb_lines = [ line.rstrip('\n') for line in
                 traceback.format_exception(ex.__class__, ex, ex_traceback)]
    exception_logger.log(tb_lines)

Here's how you can use it in Python 3:

try:
    x = get_number()
except Exception as ex:
    log_traceback(ex)

1 answer to this question.

0 votes

Hii,

In most applications, you won't be calling logging.exception(e) directly. Most likely you have defined a custom logs specific for your application or module like this:

# Set the name of the app or module
my_logger = logging.getLogger('NEM Sequencer')
# Set the log level
my_logger.setLevel(logging.INFO)

# Let's say we want to be fancy and log to a graylog2 log server
graylog_handler = graypy.GELFHandler('some_server_ip', 12201)
graylog_handler.setLevel(logging.INFO)
my_logger.addHandler(graylog_handler)

In this case, just use the logger to call the exception(e) like this:

try:
    1/0
except ZeroDivisionError, e:
    my_logger.exception(e)
answered May 11, 2020 by Niroj
• 82,880 points

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