Replace default handler of Python logger

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I've got the following code running on each request of a wsgi (web2py) application:

import logging, logging.handlers
from logging import StreamHandler, Formatter

def get_configured_logger(name):

    logger = logging.getLogger(name)

    if (len(logger.handlers) == 0):
        # This logger has no handlers, so we can assume it hasn't yet been configured (Django uses similiar trick)

        # === Configure logger ===

        # Create Formatted StreamHandler:
        FORMAT = "%(process)s %(thread)s: %(message)s"
        formatter = logging.Formatter(fmt=FORMAT)
        handler = logging.StreamHandler()
        handler.setFormatter(formatter)
        logger.addHandler(handler)
        logger.setLevel(logging.DEBUG)
        logger.debug('CONFIGURING LOGGER')

    return logger

# Get app specific logger:
logger = get_configured_logger(request.application)
logger.debug("TEST")

It's meant to configure the logger once, with the formatted handler I want. It works, except that I'm getting double entries in my stdout:

81893 4329050112: CONFIGURING LOGGER
DEBUG:dummy:CONFIGURING LOGGER
81893 4329050112: TEST
DEBUG:dummy:TEST

How do I use my new formatted handler and get rid of/hide the default one?

logging python
2 Answers

You can remove the default handler from the getLogger() using this:

logging.getLogger().removeHandler(logging.getLogger().handlers[0])

Or clear the existing handlers first before adding handlers that you want:

logging.getLogger().handlers.clear()

After doing so, these logs will no longer display except the new handlers you have added:

DEBUG: Do stuff
WARNING: Do stuff
user_150520500
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