Question

I am trying to keep a somewhat organized directory structure as I plan to add more and more scripts. So lets say I have a structure like this:

/src/main.py
/src/db/<all my DB conn & table manipulation scripts>
/src/api/<all my scripts that call different APIs>

My main.py script will include certain classes from db & api folders as needed. I have the blank _____init_____.py files in each folder so they are included fine. But say I want to include a class from the db folder in a script in the api folder? Like I would need to back up one dir somehow? The api scripts fail when I have a line like this in them:

from db.Conn import QADB

I am on v2.6.

Update: I have tried a relative import, but get this?

from ..db.Conn import QADB 
     ^ SyntaxError: 
invalid syntax

Answer 1

The way you have it setup is creating three different modules – this may or may not be what you want to do. If you want a common module which can manage different tasks you could arrange it as follows:

mymodule
 |- __init__.py
 |--database
 |   |- __init__.py
 |   |- dbclasses.py
 | 
 |--api
 |   |- __init__.py
 |   |- apiclasses.py
 |
 |--other

[etc]

If you have it like this and want to use the API and database functionalities, you can start by saying:

from mymodule.database.dbclasses import MyDBClass
from mymodule.api.apiclasses import MyAPIClass

Notice the way you had it: the name of your top “module” was src (it was not a module because it did not have an __init__.py file).

If you use many common functionalities in the top module (from either submodule) you can include them in the top __init__.py and simply call from mymodule import MyDBClass, MyAPIClass.

Contents of top __init__.py:

from mymodule.database.dbclasses import MyDBClass
from mymodule.api.apiclasses import MyAPIClass
__all__ = ['MyDBClass', 'MyAPIClass']