How can I get the file name of the function that is passed to my decorator in python?
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28-06-2021 - |
Question
I want to get the original file/scriptname, etc of the function that is being decorated. How can I do that?
def decorate(fn):
def wrapped():
return "scriptname: " + fn.scriptname?
return wrapped
I tried using fn.__code__
but that gave me more stuff that I needed. I could parse that string to get the function name, but was wondering if there is a more elegant way to do it
La solution
Try this:
return "filename: " + fn.func_code.co_filename
Autres conseils
import inspect
inspect.getfile(fn)
This won't work for builtin functions though, you have to fall back to inspect.getmodule
for those.
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