Get the current path of file executing module code

Question

I have a here.py that gets the nearest folder name:

import os
def here():
    return os.path.basename(os.path.dirname(__file__))

I have app.py which imports here.py and calls it:

from this_app.somewhere import here
print(here.here())

Finally, this is my folder structure:

this_app
  ├ __init__.py
  ├ app.py
  └ somewhere
    ├ __init__.py
    └ here.py

So you can see when I run app.py it will print somewhere because it is executing code in here.py

I want to keep the code in here.py but make it print out the folder app.py is in. In otherwords, I want to print the folder of the file that imported the module instead of the folder where the module is located, even though I want the code that decides what to print to be in the module.

I hope that makes sense. Basically, how do I change here.py so that when I call app.py it prints this_app (that is, it prints the folder name of wherever app.py is located)?

(One last consideration: my actual use case is more complicated than this so it can't resort to the current working directory because I might start the app.py from a directory far away from any of this. It has to be relative to the file that imports the here module, namely app.py.)

rdas · Accepted Answer · 2019-05-07 17:37:50Z

1

The __file__ attribute is defined per-module. You need to pass the __file__ of the calling module to here

here.py

import os
def here(calling_path):
    return os.path.basename(os.path.dirname(calling_path))

And use it:

from this_app.somewhere import here
print(here.here(__file__))

answered May 7, 2019 at 17:37

rdas

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