How to get the project directory in pyinvoke?

Viewed 38

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?

0 Answers
Related