how to get the caller's filename, method name in python

Question

for example, a.boo method calls b.foo method. In b.foo method, how can I get a\'s file name (I don\'t want to pass __file__

Answer 1

This can be done with the inspect module, specifically inspect.stack:

import inspect
import os.path

def get_caller_filepath():
    # get the caller's stack frame and extract its file path
    frame_info = inspect.stack()[1]
    filepath = frame_info[1]  # in python 3.5+, you can use frame_info.filename
    del frame_info  # drop the reference to the stack frame to avoid reference cycles

    # make the path absolute (optional)
    filepath = os.path.abspath(filepath)
    return filepath

Demonstration:

import b

print(b.get_caller_filepath())
# output: D:\Users\Aran-Fey\a.py