gristlabs/grist-core

Implement __dict__ for Record objects

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#932 创建于 2024年4月9日

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 (3 评论) (0 反应) (1 负责人)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 🙂

贡献者指南

技术栈
pythontypescript
领域
databackend
议题类型
feature
难度面向新贡献者的预计实现难度，1 表示很小改动，5 表示专家级工作。
3
预计时间有经验贡献者完成调查、实现、测试并准备 pull request 的粗略时间范围。
under 1 hour
活动状态议题当前的可参与程度：新鲜、活跃、陈旧、阻塞或等待维护者输入。
blocked
清晰度议题是否清楚说明期望改动、验收标准和下一步。
clear
前置要求
Python dict methodGrist Record class location
新手友好度1-100 的估计分数，表示该议题对首次贡献者的友好程度。
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.
Implement __dict__ for Record objects · gristlabs/grist-core#932 | Good First Issue