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fix(db, db-ivm): support Temporal objects in join hashing and normalization #1370
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5dd91ac
fix(db-ivm): hash Temporal objects by value instead of identity
goatrenterguy 962ed2f
fix(db, db-ivm): support Temporal objects in join hashing and normali…
KyleAMathews 7b6add2
Merge PR #1368 from goatrenterguy (Ben Guericke)
KyleAMathews bfea297
fix: replace as-any casts with proper TemporalLike types
KyleAMathews bcf3ef5
fix: use real Temporal objects in hash tests instead of mocks
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,6 @@ | ||
| --- | ||
| '@tanstack/db': patch | ||
| '@tanstack/db-ivm': patch | ||
| --- | ||
|
|
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| Fix Temporal objects breaking live query updates when used with joins. Temporal objects (e.g. `Temporal.PlainDate`) have no enumerable properties, so the structural hash function produced identical hashes for all Temporal values, causing join index updates to be silently swallowed. Also add Temporal support to value normalization for join key matching and to the comparator for correct sort ordering. |
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Use Temporal compare semantics instead of raw
toString()ordering.Line 58-Line 65 can misorder chronological values (e.g.,
Temporal.ZonedDateTimeacross different zones/offsets). It also doesn’t enforce same-type comparison despite the comment.💡 Proposed fix
// If both are Temporal objects of the same type, compare by string representation if (isTemporal(a) && isTemporal(b)) { - const aStr = a.toString() - const bStr = b.toString() - if (aStr < bStr) return -1 - if (aStr > bStr) return 1 - return 0 + const aTag = a[Symbol.toStringTag] + const bTag = b[Symbol.toStringTag] + + // Prefer Temporal's native compare when both values share the same constructor/type. + if (aTag === bTag && a.constructor === b.constructor) { + const compare = (a.constructor as { compare?: (x: unknown, y: unknown) => number }).compare + if (typeof compare === `function`) { + return compare(a, b) + } + } + + // Deterministic fallback for mixed Temporal types or missing compare(). + if (aTag < bTag) return -1 + if (aTag > bTag) return 1 + const aStr = a.toString() + const bStr = b.toString() + if (aStr < bStr) return -1 + if (aStr > bStr) return 1 + return 0 }🤖 Prompt for AI Agents