OpenFn / OpenFn/kit

When executing a workflow through a project, output to worklflow/tmp

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Dominant language
TypeScript
Stars
21
Forks
23
Avg merge
1d 22h
Merged PRs (30d)
17

Description

Users can execute a workflow with the command:

openfn my-workflow

This will load the current project and work out where my-workflow is from there.

When we do this, by default the output path will be my-workflow/output.json . Which is odd because the workflow lives in workflows/my-workflow. It means a messy new diff has been created and its not a nice experience.

It would be nicer to default the output to tmp/output.json, relative to the workflow.yaml file.

What about input paths? If I do -s input.hson should that be relattve to workflow/tmp? cwd? Perhaps if you do ./state.json we'll do it relative to the workspace root, but if you just do state.json we load it from /tmp. Then again its not hard to do -s tmp/input/json.

So what if we say that all input and output paths for a workflow are relative to the workflow yaml? If no path is included we do tmp/state.

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First steps

  1. Read the whole issue, then the project's contributing guide.
  2. Comment on the issue to say you are picking it up — it saves two people doing the same work.
  3. Fork the repository and make your change on a branch.
  4. Open a pull request that references the issue number.

Research direction

Start at the openfn my-workflow command and trace how it resolves workflow.yaml and selects the default output path. Review the existing -s input-path handling, then settle the relative-path rules before changing behavior; done means input and output paths consistently follow the agreed workflow-relative defaults.

Written by the indexing model from the issue text.

Assessment

Tech stack
typescript
Domain
cli
Issue type
Feature
Difficulty
4/5
Estimated time
3-5 days
Activity status
Quiet
Clarity
Needs clarification
Newbie friendliness
45/100

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