python - Automatic selection of file handler depending on file extension -

this how see it:

class basehandler:     def open(self): pass     def close(self): pass  class txthandler(basehandler):     def open(self): pass     def close(self): pass  class xmlhandler(basehandler):     def open(self): pass     def close(self): pass  def open_file(file_path):     handler = basehandler(file_path) 

for example, if file_path '..\file.xml' must return xmlhandler. please tell me, need implement functionality?

i know can implement via if-elif-else statement, i'm trying avoid dozen elif.

this preferred pythonic way:

import os handlers = {'.xml': xmlhandler, '.txt': txthandler}  def open_file(file_path):     ext = os.path.splitext(file_path)[1]     handler = handlers.get(ext, basehandler) 

in above code associate handlers extensions using dictionary.

in open_file function extract extension , use handler dictionary, considering case key doesn't exist.

i so:

if ext in handlers:     handler = handlers[ext] else:     handler = basehandler 

but of course using get method of dictionary nicer!