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解决轮询问题

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leslie
2025-07-25 20:27:20 +08:00
parent fe1300de65
commit 4325de90e1
1 changed files with 8 additions and 8 deletions
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16
src/services/claudeAccountService.js
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@@ -444,11 +444,11 @@ class ClaudeAccountService {
}
// 如果没有映射或映射无效，选择新账户
// 优先选择最近刷新过token的账户
// 优先选择最久未使用的账户（负载均衡）
const sortedAccounts = activeAccounts.sort((a, b) => {
const aLastRefresh = new Date(a.lastRefreshAt || 0).getTime();
const bLastRefresh = new Date(b.lastRefreshAt || 0).getTime();
return bLastRefresh - aLastRefresh;
const aLastUsed = new Date(a.lastUsedAt || 0).getTime();
const bLastUsed = new Date(b.lastUsedAt || 0).getTime();
return aLastUsed - bLastUsed; // 最久未使用的优先
});
const selectedAccountId = sortedAccounts[0].id;
@@ -544,11 +544,11 @@ class ClaudeAccountService {
return aRateLimitedAt - bRateLimitedAt; // 最早限流的优先
});
} else {
// 非限流账户按最近刷新时间排序
// 非限流账户按最后使用时间排序（最久未使用的优先）
candidateAccounts = candidateAccounts.sort((a, b) => {
const aLastRefresh = new Date(a.lastRefreshAt || 0).getTime();
const bLastRefresh = new Date(b.lastRefreshAt || 0).getTime();
return bLastRefresh - aLastRefresh;
const aLastUsed = new Date(a.lastUsedAt || 0).getTime();
const bLastUsed = new Date(b.lastUsedAt || 0).getTime();
return aLastUsed - bLastUsed; // 最久未使用的优先
});
}