fix(release-cleanup): selectReleasesToDelete 在時間相同時以 id 決定順序
排序加入次要鍵(id 數值,大者視為新),避免多筆 created_at 相同時保留/ 刪除的選擇因引擎排序穩定性而不確定。 Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>
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co-authored by
Claude Opus 4.8
parent
e0ab6f2693
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9e33cb8ba8
+7
-2
@@ -18,9 +18,14 @@ export function selectReleasesToDelete(releases, keepCount) {
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return Number.isNaN(time) ? 0 : time
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return Number.isNaN(time) ? 0 : time
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}
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}
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// 先一次性計算每筆的時間戳,避免在排序比較中重複呼叫 Date.parse。
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// 先一次性計算每筆的時間戳,避免在排序比較中重複呼叫 Date.parse。
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// 時間相同時以 id(數值,大者為新)作為次要鍵,確保排序結果具決定性。
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return releases
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return releases
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.map((release) => ({ release, time: createdTime(release) }))
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.map((release) => ({
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.sort((a, b) => b.time - a.time)
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release,
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time: createdTime(release),
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id: Number(release?.id) || 0,
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}))
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.sort((a, b) => b.time - a.time || b.id - a.id)
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.map((entry) => entry.release)
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.map((entry) => entry.release)
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.slice(keepCount)
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.slice(keepCount)
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}
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}
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