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In my pyinvoke task library, I would like to know the project directory so I can derive default locations for the directories like src, test, and dist.

Currently, the only way I can see to do this is from context.config._project_prefix.

Is there a better way to get the project directory? One that doesn't require the use of an underscore-prefixed (private, subject to change) attribute? Alternatively, what's the reason for why no such public attribute exists?

davidrmcharles
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1 Answers1

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What I've often done is to use __file__ in my tasks.py and find the project root relative to that.

For example:

myproject
    dist
    src
    test
    tasks.py

# In tasks.py

from pathlib import Path

project_dir = Path(__file__).parent.absolute()
Eric Smith
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