yuming
d8573359a2
atRisk.js 之前只按 fireInDays 排序，同一天内多条提醒的先后完全未定义， 和 reminder.js 实际发送时「当天>提前、重要程度降序」的规则对不上，导致 额度卡在中间时首页提示的风险人名和定时任务实际跳过的人不一致。 - 把 importanceRank 从 reminder.js 抽到共享的 src/importance.js，两处复用 - atRisk.js 的 computeAtRisk 排序改为三段式：fireInDays 升序 → 同天内 kind（当天优先于提前）→ 同 kind 内重要程度降序 - 补充 atRisk.test.js 用例覆盖「同一天多条事件、额度只够一部分」场景
126 lines
5.1 KiB
JavaScript
126 lines
5.1 KiB
JavaScript
const test = require('node:test')
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const assert = require('node:assert')
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const { expandEvents, computeAtRisk, LOOKAHEAD_DAYS } = require('../src/atRisk')
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const TODAY = new Date(2026, 7, 15) // 2026-08-15
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// 距今 10 天的公历纪念日，提前 3 天提醒
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const near = {
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id: 'a1', personName: '张三', remindEnabled: 1, importance: 'high',
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isLunar: false, solarMonth: 8, solarDay: 25, remindDays: 3
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}
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// 距今 40 天
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const mid = {
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id: 'a2', personName: '李四', remindEnabled: 1, importance: 'low',
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isLunar: false, solarMonth: 9, solarDay: 24, remindDays: 7
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}
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test('LOOKAHEAD_DAYS 为 60', () => {
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assert.strictEqual(LOOKAHEAD_DAYS, 60)
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})
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test('一条纪念日展开出「当天」和「提前」两个事件', () => {
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const events = expandEvents(near, TODAY)
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assert.strictEqual(events.length, 2)
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const onDay = events.find(e => e.kind === 'onDay')
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const ahead = events.find(e => e.kind === 'ahead')
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assert.strictEqual(onDay.fireInDays, 10)
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assert.strictEqual(ahead.fireInDays, 7)
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})
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test('remindDays 为 0 时只有当天事件', () => {
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const events = expandEvents({ ...near, remindDays: 0 }, TODAY)
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assert.strictEqual(events.length, 1)
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assert.strictEqual(events[0].kind, 'onDay')
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})
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test('超出 60 天窗口的事件被过滤掉', () => {
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const far = { ...near, solarMonth: 12, solarDay: 25, remindDays: 3 }
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assert.strictEqual(expandEvents(far, TODAY).length, 0)
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})
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test('提前日已过、当天未到时，只剩当天事件', () => {
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// 距今 2 天，但提前 7 天的时点早已过去
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const soon = { ...near, solarMonth: 8, solarDay: 17, remindDays: 7 }
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const events = expandEvents(soon, TODAY)
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assert.strictEqual(events.length, 1)
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assert.strictEqual(events[0].kind, 'onDay')
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})
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test('余额足够时没有风险', () => {
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const r = computeAtRisk([near, mid], 10, TODAY)
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assert.strictEqual(r.atRiskCount, 0)
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assert.deepStrictEqual(r.atRiskNames, [])
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})
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test('余额为 0 时全部有风险', () => {
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const r = computeAtRisk([near, mid], 0, TODAY)
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assert.strictEqual(r.atRiskCount, 4)
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assert.deepStrictEqual(r.atRiskNames, ['张三', '李四'])
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})
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test('按时间先后消耗额度，后面的才有风险', () => {
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// 张三的两个事件在第 7、10 天，李四的在第 33、40 天
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const r = computeAtRisk([near, mid], 2, TODAY)
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assert.strictEqual(r.atRiskCount, 2)
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assert.deepStrictEqual(r.atRiskNames, ['李四'])
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})
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test('人名去重，同一人占两条只出现一次', () => {
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const r = computeAtRisk([near], 0, TODAY)
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assert.strictEqual(r.atRiskCount, 2)
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assert.deepStrictEqual(r.atRiskNames, ['张三'])
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})
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test('未开启提醒的纪念日不参与计算', () => {
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const r = computeAtRisk([{ ...near, remindEnabled: 0 }], 0, TODAY)
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assert.strictEqual(r.atRiskCount, 0)
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})
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// ------ 同一天内排序需与 reminder.js 的 runForUser 完全对齐 ------
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// 四条纪念日都恰好在第 5 天产生一个事件：
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// 甲(当天/high) 乙(当天/low) 丙(提前/high) 丁(提前/medium)
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// 丙、丁的「当天」事件被安排在第 65 天（超出 60 天窗口），不会进入计算，
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// 这样每人恰好只贡献一个事件，排序结果可以精确断言。
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const onDayHigh = {
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id: 'e1', personName: '甲', remindEnabled: 1, importance: 'high',
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isLunar: false, solarMonth: 8, solarDay: 20, remindDays: 0
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}
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const onDayLow = {
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id: 'e2', personName: '乙', remindEnabled: 1, importance: 'low',
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isLunar: false, solarMonth: 8, solarDay: 20, remindDays: 0
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}
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const aheadHigh = {
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id: 'e3', personName: '丙', remindEnabled: 1, importance: 'high',
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isLunar: false, solarMonth: 10, solarDay: 19, remindDays: 60
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}
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const aheadMedium = {
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id: 'e4', personName: '丁', remindEnabled: 1, importance: 'medium',
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isLunar: false, solarMonth: 10, solarDay: 19, remindDays: 60
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}
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// 刻意打乱输入顺序（不按「当天>提前、重要程度降序」排列），
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// 这样如果排序逻辑退化成只按 fireInDays 排（Array.sort 是稳定排序，
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// 同值会保留输入顺序），测试才能真正暴露出排序规则失效，而不是被输入顺序碰巧掩盖。
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const sameDayGroup = [aheadMedium, aheadHigh, onDayLow, onDayHigh]
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test('同一天内，当天事件排在提前事件之前（与 reminder.js 一致）', () => {
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// 额度只够 2 条：应先消耗「当天」的甲、乙，风险留给「提前」的丙、丁
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const r = computeAtRisk(sameDayGroup, 2, TODAY)
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assert.strictEqual(r.atRiskCount, 2)
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assert.deepStrictEqual(r.atRiskNames, ['丙', '丁'])
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})
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test('同一天内，当天事件按重要程度降序（high 先于 low）', () => {
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// 额度只够 1 条：当天里 high 的甲应排在 low 的乙前面，先被消耗
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const r = computeAtRisk(sameDayGroup, 1, TODAY)
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assert.strictEqual(r.atRiskCount, 3)
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assert.deepStrictEqual(r.atRiskNames, ['乙', '丙', '丁'])
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})
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test('同一天内，提前事件按重要程度降序（high 先于 medium）', () => {
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// 额度够 3 条：甲、乙（当天）先消耗，提前事件里 high 的丙应先于 medium 的丁被消耗
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const r = computeAtRisk(sameDayGroup, 3, TODAY)
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assert.strictEqual(r.atRiskCount, 1)
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assert.deepStrictEqual(r.atRiskNames, ['丁'])
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})
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