CODEWITHSUNDEEP
pythonlogging
JS.
JS.

Reputation: 16097

Configuring child loggers

Every time I think I understand the logging module, gremlins come in and change the way it works. (Ok, I'll admit, that gremlin may be me changing my code.)

What am I doing wrong here?

> ipython
> import logging
> log = logging.Logger("base")
> log.addHandler(logging.StreamHandler())

> log.critical("Hi")
Hi

> log2 = log.getChild("ment")

> log2.critical("hi")
No handlers could be found for logger "base.ment"

I could have sworn that in the past, I was able to use child loggers without additional configuration...

Upvotes: 13

Views: 15675

Answers (2)

Silas Ray
Silas Ray

Reputation: 26150

More detail: You are using the module wrong. :) From looking at the module code, it looks like they don't expect you to ever create a logging.Logger() directly. Many of the functions available directly on the module (ex getLogger()) and the methods on logging.Logger() (ex getChild()) actually proxy through an instance of logging.Manager that the module creates on import. When you create a Logger with logging.Logger() directly, you are actually creating a Logger instance outside of the Manager. When you subsequently call log.getChild(), the module is actually creating the new logger inside the Manager, but with the name of the Manager-external logger appended to the front of the logger name. So your handler added to log is not in the Manager with the spawned child, and thus the handler doesn't work. I am a little confused still though on why adding a handler to log before or after creating log2 causes logging against log2 to behave differently. I don't see what's causing that...

Upvotes: 3

unutbu
unutbu

Reputation: 879331

If you change 

log = logging.Logger('base')

to

log = logging.getLogger('base')

then it works:

import logging

log = logging.getLogger('base')
log.addHandler(logging.StreamHandler())
log.critical('Hi')
log2 = log.getChild('ment')
log2.critical('hi')

yields

Hi
hi

Upvotes: 13

Related Questions