Sort same ranked items alphabetically #64
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kentcdodds
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Jun 6, 2019
| const same = aRank === bRank | ||
| if (same) { | ||
| if (aKeyIndex === bKeyIndex) { | ||
| return aIndex < bIndex ? aFirst : bFirst | ||
| return aRankedItem.localeCompare(bRankedItem) |
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Is there ever a case when aRankedItem or bRankedItem could be anything but a string?
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Tough to think through, the tests cover every situation i thought of.
- No matches
- Null values
- Array of values
- Array of objects with key
- Nested objects with keys
- Array of objects with multiple keys
- Array of objects with key that points to array
An option as a "just-in-case" would be to add a typeof check on the xRankItems and add the indexes back in as a else clause but it would not meet 100% test branch coverage.
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I think we're probably ok. Fixing any issues would be simple and quick so we'll go with this. Thanks!
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What:
Why:
Currently, if items are the same rank with the same key based index, the items are sorted based on their index in the starting input. This is fine in general but I believe it makes more sense to sort them alphabetically when they have the same ranking and also by length allowing the smaller words to float to the top and make filtering a progressive act.
Example:
search:
'ap'input:
['wrapped','snap', 'pap']old output:
['wrapped','snap', 'pap']new output:
['pap', 'snap', 'wrapped']How:
We can take advantage of the base String.prototype.localeCompare() to do a comparison while still accepting diacritics.
Checklist:
Comments
This may need discussion and you may not like the approach or the idea at all, just let me know or if you find any issues.
Thanks!