gristlabs/grist-core

Implement __dict__ for Record objects

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#932 建立於 2024年4月9日

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 (3 留言) (0 反應) (1 負責人)TypeScript (3,613 star) (168 fork)batch import
good first issuegouv.frpython

描述

Hey there 👋🏼

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 🙂

貢獻者指南

技術棧
pythontypescript
領域
databackend
議題類型
feature
難度面向新貢獻者的預計實作難度，1 表示很小改動，5 表示專家級工作。
3
預計時間有經驗貢獻者完成調查、實作、測試並準備 pull request 的粗略時間範圍。
under 1 hour
活動狀態議題目前的可參與程度：新鮮、活躍、陳舊、阻塞或等待維護者輸入。
blocked
清晰度議題是否清楚說明預期改動、驗收標準和下一步。
clear
前置要求
Python dict methodGrist Record class location
新手友善度1-100 的估計分數，表示該議題對首次貢獻者的友善程度。
70
研究方向
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.