CODEWITHSUNDEEP
pythonpython-2.7
JavaSa
JavaSa

Reputation: 6241

Getting entry point scripts path in scripts called by it

Consider you have have a script called /home/user/pack/test.py

Within this script we call some helper class which is defined in other file path(using import)
Lets say the Helper file path is: /home/user/pack/abc/helper.py

I wonder if there is a way to get the original launching script path /home/user/src/test.py within the Helper script for some usage.

I tried using __file__ but it gave me /home/user/pack/abc/helper.py of course.

Note: I don't want to pass the launching script path(__file__) as parameter to the Helper class.

Upvotes: 2

Views: 514

Answers (1)

Hayk Davtyan
Hayk Davtyan

Reputation: 917

You can use the inspect module from the Standard Library to achieve your goal, this way:

helper.py

def some_helper():
    print(inspect.stack()[1][1])

test.py

import helper

helper.some_helper()

So by executing test.py you will get "path/to/test.py" as your output inside the helper.py function.

Upvotes: 3

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