Python Module import 3.7

Question

Hey I have a tree structure with an empty init.py in each directory/subdirectory. However, I get a failed to load module Utilities from my Home directory .py file.

Using python 3.7

My tree looks like the following:

 C:.
├───Tests
│ ├───Checkout
│ ├───GlobalFooter
│ ├───GlobalHeader
│ │ └───__pycache__
│ ├───Home
│ │ └───__pycache__
│ ├───MyAccount
│ ├───ProductDetail
│ ├───ProductResults
│ │ └───__pycache__
│ └───SignIn
└───Utilities
 └───__pycache__

I have tried the following:

sys.path.insert(0, 'C:/Web2/TSC.WebFactory.Web2.Tests/Utilities')
from Utilities.utils import addCookies, configureOptions

Answer 1

If you want to access your whole tree of modules by their package qualified names, you shouldn't include Utilities in the sys.path entry; from Utilities.utils import ... assumes that some folder in sys.path has a package/folder named Utilities, that contains either a subpackage/folder named utils or a submodule/file named utils.py; by adding 'C:/Web2/TSC.WebFactory.Web2.Tests/Utilities' to sys.path, it expects 'C:/Web2/TSC.WebFactory.Web2.Tests/Utilities/Utilities/utils.py' (note doubled Utilities).

The solution here is to remove that final directory from the path:

sys.path.insert(0, 'C:/Web2/TSC.WebFactory.Web2.Tests')

Now from Utilities.utils import ... will look in C:\Web2\TSC.WebFactory.Web2.Tests for Utilities\utils.py and find it as expected.

Side-note: If you want to keep to Windows standard backslashes as directory separators, you can use them fairly cleanly by just making the path a raw string literal, avoiding the need for constant escaping:

sys.path.insert(0, r'C:\Web2\TSC.WebFactory.Web2.Tests')

Answer 2

If you haven't already tried, do this as when doing sys.path.insert you shouldn't have to specify the folder on the import line.

from utils import addCookies, configureOptions