fix: TypeScript null check for originalUserId
Co-Authored-By: Claude Opus 4.5 <noreply@anthropic.com>
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@ -587,7 +587,9 @@ export class UsersService {
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});
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const userIdToSeq = new Map(
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ancestorReferrals.map((r) => [r.originalUserId.toString(), r.accountSequence]),
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ancestorReferrals
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.filter((r) => r.originalUserId !== null)
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.map((r) => [r.originalUserId!.toString(), r.accountSequence]),
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);
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// 按顺序获取每个祖先的详细信息
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