使用三个版本(类、函数、lambda)来设置日志记录过滤器

Question

>我可以通过三种不同的方式创建相同的过滤器：

1(作为λ：

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

2(作为功能：

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(作为一个类

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，过滤器是否执行以下操作：

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

或者它如何"知道"是调用函数还是调用其中一个类方法？

Answer 1

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

以下是源代码：

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