Getting path of originally executed script in python?

Question

I have the following working directory from which I am running ankiAdd.py - in this script at one point i'm accessing the 'internal API' of another python program elsewhere on my hard drive in order to use some of those functions. I would like to get the working directory where 'ankiAdd.py' is stored however I kept retrieving another directory.

The following snippet is a function within my ankiAdd.py script. Collection is imported from another directory on my hard drive in which another program Anki is installed. I'd like the file_pattern variable to give the directory of this script and not the directory of cpath, which is currently happening.

sys.path.append("C:/Users/Shaun/Documents/Personal_Projects/ankiExperiment/anki-scripting/anki")
from anki.storage import Collection
PROFILE_HOME = os.path.expanduser("C:/Users/Shaun/Desktop/User 1")
cpath = os.path.join(PROFILE_HOME, "collection.anki2")

@eel.expose
def callLoad():
    col = Collection(cpath, log=True)
    json_pattern = "word_data.json"
    file_pattern = os.path.join(os.getcwd(),json_pattern)
    load(col,file_pattern)

I've tried numerous things like os.getcwd() and os.path.dirname(os.path.realpath(file)) however I kept ending up retrieving C:\Users\User\Desktop\User 1\collection.media which is the location of cpath. How can I get the directory of ankiAdd.py which is the script which was originally run?

edit:

def callLoad():
    col = Collection(cpath, log=True)
    json_pattern = "word_data.json"
    print(os.path.dirname(os.path.realpath(__file__)))
    file_pattern = os.path.join(os.getcwd(),json_pattern)
    #load(col,file_pattern)

if __name__ == '__main__':
callLoad()

if I run this code, the directory returned is C:\Users\Shaun\Desktop\User 1\collection.media which is the directory set by this at the top of my file:

PROFILE_HOME = os.path.expanduser("C:/Users/Shaun/Desktop/User 1")

I am running the script from

inim · Accepted Answer · 2019-12-19 19:31:52Z

1

If I understand your question correctly, you're trying to get the location of a file or module you're importing, and keep getting the location of the file or module that's doing the import. This can be resolved!

Anything you import will have a __file__ attribute. If you had, for instance,

import anki

print(anki.__file__)

you would get the location of the anki module's __init__.py file (or if anki is just a Python file, you'd get the location of the file itself).

answered Dec 19, 2019 at 19:31

inim

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1 Comment

Blargian Over a year ago

I'm actually trying to get the location of the .py script in my working directory and using __ file __ I'm getting back this path which I set as a variable PROFILE_HOME = os.path.expanduser("C:/Users/Shaun/Desktop/User 1"). I'd like to get the location of the file from which I call import anki

KPandian · Accepted Answer · 2019-12-19 19:47:00Z

0

file returns the location of the script in execution. You can try this.

print(__file__)

answered Dec 19, 2019 at 19:47

KPandian

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3 Comments

Blargian Over a year ago

If I call a function callLoad() from if name == 'main' where I have print(os.path.dirname(os.path.realpath(file))) inside of that function the directory that is returned is not the directory of the .py script that I am running. (Please see my edit)

KPandian Over a year ago

Can you try this in place of file and share me the output path ? " print(sys.argv[0]) " Note: Please do import "sys" and "os" packages.

Blargian Over a year ago

Thank you. sys and os are both imported. print(sys.argv[0]) in that function gives back "ankAdd.py" - is there a way to display the full path?

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