如何删除日志过滤器/格式器

问题描述 投票:0回答:2

我有以下日志记录过滤器:

class LogFilter:
    """
    Print all messages that do not equal hello."""
    def filter(self, record):
        print ('@@@@@@@ Filter', record)
        if getattr(record, 'msg') == 'Hello':
            return False
        return True
    def __repr__(self):
        return 'LogFilter'

然后:

import logging
handler = logging.StreamHandler()
handler.addFilter(LogFilter())
logger = logging.getLogger(__name__)
logger.addHandler(handler)

我将如何删除此过滤器?例如,当我尝试:

>>> handler.removeFilter(LogFilter())
>>> handler.removeFilter(LogFilter)
>>> handler.removeFilter('LogFilter') # ????
>>> handler.filters
[LogFilter] # filter is still there.

我还有过滤器。删除过滤器(或格式化程序)的正确方法是什么?

python python-3.x logging
2个回答
1
投票

如果您查看removeFilter的源代码

def removeFilter(self, filter):
    """
    Remove the specified filter from this handler.
    """
    if filter in self.filters:
        self.filters.remove(filter)

您看到您需要实际的filter对象才能删除它。一个快速的解决办法是

import logging
handler = logging.StreamHandler()
LF = LogFilter()
handler.addFilter(LF)
logger = logging.getLogger(__name__)
logger.addHandler(handler)

然后,当您要删除过滤器时

>>> handler.removeFilter(LF)
>>> handler.filters
[]

0
投票

将过滤器作为对象传递,为什么由于列表中的handler.filters和当您正在执行handler.removeFilter('name')而又是handler.filters.remove('name')时却引起问题呢?>

列表无法删除过滤器值并创建值错误。因此LogFIlter停留在列表中,但是如果将filter作为对象传递并将其删除,则它可以工作

import logging
handler = logging.StreamHandler()
log_filter  = LogFilter()
handler.addFilter(log_filter)
logger = logging.getLogger(__name__)
logger.addHandler(handler)
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