New submission from Sean <sean.davi...@googlemail.com>:
Using "logging" module I can do the following:
f = logging.Formatter("%(levelname)s: %(name)s: %(message)s")
h = logging.StreamHandler()
h.setLevel(logging.NOTSET)
h.setFormatter(f)
l = logging.getLogger()
l.addHandler(h)
l2 = logging.getLogger('x.y.z')
l2.setLevel(logging.INFO)
print len(l2.handlers)
l2.critical('critical')
l2.error('error')
l2.warning('warning')
l2.info('info')
l2.debug('debug')
Note, the l2(x.y.z) logger does not have a handler (it just propagates
to the l(root) logger's handler) but the l2 logger does have a different
level set.
When I try to do this with logging.config.fileConfig it fails saying I
have not specified a handler for the x.y.z logger. But I don't want to
specify a handler, I just want to set the level and have it propagate up
the hierarchy.
Am I doing something wrong?
Sean.
P.S. Here is the config file:
[loggers]
keys=root,xyz
[handlers]
keys=consoleStderr
[formatters]
keys=simpleFormatter
[logger_root]
level=NOTSET
handlers=consoleStderr
[logger_xyz]
level=DEBUG
qualname=x.y.z
[handler_consoleStderr]
class=StreamHandler
level=NOTSET
formatter=simpleFormatter
args=(sys.stderr,)
[formatter_simpleFormatter]
%(levelname)s: %(name)s: %(message)s
----------
messages: 90392
nosy: sean
severity: normal
status: open
title: With logging.config.fileConfig can't create logger without handler
type: behavior
versions: Python 2.6
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue6458>
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