Why does this python logging function not work?

Question

For some reason, the code below generates an empty log file, with no text inside. What am I doing wrong?

import logging

def log_generator(logfilepath='test.log'):

    my_formatter = logging.Formatter('%(asctime)s %(levelname)s %(message)s')
    my_logger = logging.getLogger('example_logger')
    my_handler = logging.FileHandler( logfilepath )
    my_handler.setFormatter( my_formatter )
    my_logger.addHandler( my_handler )

    my_logger.info('message 1') # This should be visible in the log file


log_generator() # Execute the log_generator() function

Vor · Accepted Answer · 2014-08-07 14:26:12Z

3

You need to set level to the logger.

add this line befoure calling info():

my_logger.setLevel(logging.DEBUG)

So your code should look like this:

import logging

def log_generator(logfilepath='test.log'):

    my_formatter = logging.Formatter('%(asctime)s %(levelname)s %(message)s')
    my_logger = logging.getLogger('example_logger')
    my_handler = logging.FileHandler(logfilepath)
    my_handler.setFormatter( my_formatter )
    my_logger.addHandler( my_handler )
    my_logger.setLevel(logging.DEBUG)
    my_logger.info('message 1') # This should be visible in the log file


log_generator()

Which will five you the output:

john-the-ripper@john-the-ripper:/tmp$ more test.log 
2014-08-07 10:20:23,665 INFO message 1

Updated:

From the docs:

 The INFO message doesn’t appear because the default level is WARNING.

edited Aug 7, 2014 at 14:26

answered Aug 7, 2014 at 14:21

Vor

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2 Comments

James Mertz Over a year ago

I thought the default level was INFO on up. Is this not true?

Vor Over a year ago

@KronoS default is warning

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