fix: Better filtering by ref values #2563
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jamie-rasmussen
approved these changes
Oct 2, 2024
| if (columnIsRefExpanded(field)) { | ||
| // In this case, we actually just want to filter by the parent ref itself. | ||
| // This means we need to: | ||
| // 1. Determine the column of the highest level anscestor column with a ref |
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anscestor => ancestor
| // 1. Determine the column of the highest level anscestor column with a ref | ||
| // 2. Get the value of that corresponding cell (ref column @ row) | ||
| // 3. Add a filter for that ref on that column. | ||
| // The aknoweldged drawback of this approach is that we are not filtering by that |
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aknoweldged => acknowledged
| let ancestorField: string | null = null; | ||
| let targetRef: string | null = null; | ||
| for (let i = 0; i <= fieldParts.length; i++) { | ||
| const ancestorFieldCandidate = fieldParts.slice(0, i).join('.'); |
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slice(0,0) will be an empty list, I think we don't need to check that case?
| @@ -306,6 +373,18 @@ export const CallsTable: FC<{ | |||
| onAddFilter | |||
| ); | |||
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| // This contains columns which are suitable for selection and raw data | |||
| // entry. Noteably, not children of expanded refs. | |||
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Noteably => Notably
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Filtering by values within a reffed object is technically quite hard. However, Scott made the great recommendation today to have the following rules:
This Pr does those two things (: