替换Python logg的默认处理程序

Question

我已经在wsgi（web2py）应用程序的每个请求上运行了以下代码：

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")

这意味着配置一次记录器，使用我想要的格式化处理程序。它可以工作，除了我的标准输出中有两个条目：

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

如何使用新的格式化处理程序并删除/隐藏默认处理程序？

Answer 1

也许下面的例子会有帮助。基本上，您可以删除要禁用的记录器的处理程序，或者不与要从中进行日志记录的记录器一起传播。

$ cat testlog.py
import logging
logging.basicConfig(filename='foo', level=logging.DEBUG)
root_logger = logging.getLogger()
root_logger.debug('bar')

my_logger = logging.getLogger('my_logger')
FORMAT = "%(process)s %(thread)s: %(message)s"
formatter = logging.Formatter(fmt=FORMAT)
handler = logging.StreamHandler()
handler.setFormatter(formatter)

my_logger.addHandler(handler)
my_logger.setLevel(logging.DEBUG)
my_logger.info('baz')

my_logger.propagate = False
my_logger.info('foobar')

my_logger.propagate = True
my_logger.info('foobaz')
root_logger.handlers = []
my_logger.info('barbaz')

$ python testlog.py
5927 140735224465760: baz
5927 140735224465760: foobar
5927 140735224465760: foobaz
5927 140735224465760: barbaz

$ cat foo
DEBUG:root:bar
INFO:my_logger:baz
INFO:my_logger:foobaz