使用三个版本（class、函数、lambda）来设置日志过滤器

Question

我可以用三种不同的方式创建相同的过滤器：

1) 作为 lambda：

import logging
handler = logging.StreamHandler()
handler.addFilter(lambda record: getattr(record, 'msg') == 'Hello')
logger = logging.getLogger(__name__)
logger.addHandler(handler)
logger.warning('ok!')

2) 作为函数：

def is_hello(record):
    return getattr(record, 'msg') == 'Hello'

import logging    
handler = logging.StreamHandler()
handler.addFilter(is_hello)
logger = logging.getLogger(__name__)
logger.addHandler(handler)
logger.warning('ok!')

3) 作为 class

class LogFilter:
    def filter(self, record):
        return getattr(record, 'msg') == 'Hello'

import logging    
handler = logging.StreamHandler()
handler.addFilter(LogFilter())
logger = logging.getLogger(__name__)
logger.addHandler(handler)
logger.warning('ok!')

loggin 模块如何允许这种变化，过滤器是否会做类似的事情：

if callable(provider_filter):
    provider_filter(record)
else:
    provider_filter.filter(record)

或者'know'是调用函数还是调用class方法之一？

Answer 1

基本上，是的，它只是检查它是否具有 .filter 属性，如果没有，它假定它是可调用的：

Here is the source code:

def filter(self, record):
    """
    Determine if a record is loggable by consulting all the filters.
    The default is to allow the record to be logged; any filter can veto
    this and the record is then dropped. Returns a zero value if a record
    is to be dropped, else non-zero.
    .. versionchanged:: 3.2
       Allow filters to be just callables.
    """
    rv = True
    for f in self.filters:
        if hasattr(f, 'filter'):
            result = f.filter(record)
        else:
            result = f(record) # assume callable - will raise if not
        if not result:
            rv = False
            break
    return rv

使用三个版本（class、函数、lambda）来设置日志过滤器

Using three version (class, function, lambda) for setting logging Filter

python

python-3.x

logging

python-internals