gristlabs/grist-core

Implement __dict__ for Record objects

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#932 opened on Apr 9, 2024

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聽(3 comments)聽(0 reactions)聽(1 assignee)TypeScript聽(3,613 stars)聽(168 forks)batch import
good first issuegouv.frpython

Description

Hey there 馃憢馃徏

I now know of the RECORD function (thanks to MichaelGrist from the Discord server who kindly pointed me to it), but as a python user (and bad documentation reader) I feel like it would just make sense to be able to call dict(rec) and get a dictionary representation of the record.

Happy to implement if I get your go-ahead 馃檪

Contributor guide

Tech stack
pythontypescript
Domain
databackend
Issue type
feature
DifficultyEstimated implementation difficulty for a new contributor, from 1 for very small changes to 5 for expert-level work.
3
Estimated timeA rough time range for an experienced contributor to investigate, implement, test, and prepare a pull request.
under 1 hour
Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.
blocked
ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.
clear
Prerequisites
Python dict methodGrist Record class location
Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.
70
Research direction
Locate the Record class definition in the Grist Python codebase (likely in 'records.py'). Add a ` dict ` method that returns a dictionary of the record's fields. Ensure it works with the existing `dict()` built in. Check for existing tests for the Record class and add a test. Review the linked Discord conversation for any additional context or design decisions.
Implement __dict__ for Record objects 路 gristlabs/grist-core#932 | Good First Issue