[Django]-Naming Python loggers

by

68👍

✅

I typically don’t use or find a need for class-level loggers, but I keep my modules at a few classes at most. A simple:

import logging
LOG = logging.getLogger(__name__)

At the top of the module and subsequent:

LOG.info('Spam and eggs are tasty!')

from anywhere in the file typically gets me to where I want to be. This avoids the need for self.log all over the place, which tends to bother me from both a put-it-in-every-class perspective and makes me 5 characters closer to 79 character lines that fit.

You could always use a pseudo-class-decorator:

>>> import logging
>>> class Foo(object):
...     def __init__(self):
...             self.log.info('Meh')
... 
>>> def logged_class(cls):
...     cls.log = logging.getLogger('{0}.{1}'.format(__name__, cls.__name__))
... 
>>> logged_class(Foo)
>>> logging.basicConfig(level=logging.DEBUG)
>>> f = Foo()
INFO:__main__.Foo:Meh
👤cdleary

3👍

For class level logging, as an alternative to a pseudo-class decorator, you could use a metaclass to make the logger for you at class creation time…

import logging

class Foo(object):
    class __metaclass__(type):
        def __init__(cls, name, bases, attrs):
            type.__init__(name, bases, attrs)
            cls.log = logging.getLogger('%s.%s' % (attrs['__module__'], name))
    def __init__(self):
        self.log.info('here I am, a %s!' % type(self).__name__)

if __name__ == '__main__':
    logging.basicConfig(level=logging.DEBUG)
    foo = Foo()
👤Matt Anderson

2👍

That looks like it will work, except that self won’t have a __module__ attribute; its class will. The class-level logger call should look like:

self.log = logging.getLogger( "%s.%s" % ( self.__class__.__module__, self.__class__.__name__ ) )
👤Steve Losh

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