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
- Read the whole issue, then the project's contributing guide.
- Comment on the issue to say you are picking it up — it saves two people doing the same work.
- Fork the repository and make your change on a branch.
- 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.
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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