如何在Python中获取logging.FileHandler的文件名？

Question

A logging.FileHandler是用文件名构造的,所以有没有办法从logging.FileHandler对象中获取文件名？

我试过dir(logging.FileHandler)但没有看到任何可能的解决方案.

Answer 1

>>> import logging
>>> fh = logging.FileHandler('/Users/defuz/test.txt')
>>> fh.baseFilename
'/Users/defuz/test.txt'
>>> fh.stream.name
'/Users/defuz/test.txt'

Answer 2

请找dir(logging.FileHandler)

 ['__class__', '__delattr__', '__dict__', '__doc__', '__format__', 
  '__getattribute__', '__hash__', '__init__', '__module__', '__new__', 
  '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', 
  '__str__', '__subclasshook__', '__weakref__', '_name', '_open', 'acquire', 
  'addFilter', 'baseFilename', 'close', 'createLock', 'emit', 'encoding', 'filter', 
  'filters', 'flush', 'format', 'formatter', 'get_name', 'handle', 'handleError', 
  'level', 'lock', 'mode', 'name', 'release', 'removeFilter', 'setFormatter', 'setLevel', 
  'set_name', 'stream']

您可以选择obj.baseFilename来获取文件名.