50 lines
1.7 KiB
JavaScript
50 lines
1.7 KiB
JavaScript
/**
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* 拉取完整关联列表：e(page, limit) 返回 {list, count?}，t 判断调用上下文是否仍有效。
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* 按正整数 id 去重；总数缺失时以不足一页结束，无新增项时拒绝，避免无限分页。
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*/
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async function e(e, t = () => !0) {
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let n = [],
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r = new Set();
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for (let i = 1; ; i++) {
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if (!t()) return [];
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let a = await e(i, 100);
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if (!t()) return [];
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if (!a || !Array.isArray(a.list)) throw Error(`列表响应无效，请重试`);
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let o = a.count === void 0 ? null : Number(a.count);
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if (o !== null && (!Number.isSafeInteger(o) || o < 0))
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throw Error(`分页信息无效，请重试`);
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let s = n.length;
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for (let e of a.list) {
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if (!e || !Number.isSafeInteger(Number(e.id)) || Number(e.id) <= 0)
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throw Error(`列表项目无效，请重试`);
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r.has(String(e.id)) || (r.add(String(e.id)), n.push(e));
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}
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if (o === null ? a.list.length < 100 : n.length >= o) return n;
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if (s === n.length) throw Error(`未能加载完整列表，请刷新重试`);
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}
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}
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/** 按本次请求的每个 id 汇总批量关联结果；缺失/重复结果均不能被判作成功。 */
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function t(e, t) {
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let n = (
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Array.isArray(e) ? e : e && typeof e == `object` && `id` in e ? [e] : []
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).filter((e) => !!e && typeof e == `object`),
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r = [],
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i = [],
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a = [];
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for (let e of t) {
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let t = n.filter((t) => Number(t.id) === e);
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t.length === 1 && Number(t[0].status) === 200
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? r.push(e)
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: (i.push(e),
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a.push(
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`#${e}：${t.length === 1 ? String(t[0].msg || `关联失败`) : `未取得明确结果，请刷新核对`}`,
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));
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}
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return {
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succeeded: r,
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failed: i,
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details: a,
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};
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}
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export { e as n, t };
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