"""saver.py 的自测脚本。 用途：在没有真实 MCP Token 的情况下，验证方案枚举、排序与凑单陷阱检测的逻辑正确性。 注意：这里使用的 price_fn 是一个**本地模拟计价器**，仅用于测试算法逻辑， 绝不能用于生产环境 —— 生产环境的计价必须来自 MCP 的 calculate-price。 运行： python src/test_saver.py """ import os import sys sys.path.insert(0, os.path.dirname(os.path.abspath(__file__))) from saver import ( # noqa: E402 Coupon, Item, Plan, PricedPlan, build_base_plans, recommend, render, within_budget, ) def mock_price_fn(plan: Plan) -> PricedPlan: """模拟计价器：仅用于逻辑自测。""" goods = plan.goods_amount() discount = 0.0 if plan.coupon and plan.coupon.applicable(goods): discount = plan.coupon.discount payable = round(max(goods - discount, 0), 2) return PricedPlan(plan=plan, payable=payable, discount=discount, source="MOCK") def check_demand(mains, sides, combos, coupons) -> None: """校验需求约束在排序之前生效。""" print("=" * 60) print("2. 需求约束（demand）—— 必须在排序之前生效") print("=" * 60) def wants_burger_and_fries(plan: Plan) -> bool: """需求：麦辣鸡腿堡 + 中薯条，两样齐全。""" codes = plan.covered_codes() return "M001" in codes and "S001" in codes result = recommend( mains=mains, sides=sides, combos=combos, coupons=coupons, price_fn=mock_price_fn, demand=wants_burger_and_fries, top_k=5, ) rec = result["recommended"] # 最便宜的候选是「香芋派 8 元」——不含主食。若不先按需求筛掉， # 它会挤进推荐，用户就会得到一个「便宜但吃不饱」的答案。 assert rec, "应至少有一个满足需求的方案" assert all(wants_burger_and_fries(p.plan) for p in rec), \ "推荐里出现了不满足需求的方案 —— demand 没有在排序前生效" assert rec[0].payable > 8.0, "不满足需求的便宜方案不应胜出" print(f" {len(rec)} 个方案全部满足需求；最便宜 {rec[0].payable} 元（{rec[0].label}）") print(" [PASS] 更便宜但吃不饱的方案没有混进推荐") print() def check_combo_parts() -> None: """校验套餐构成与枚举深度。""" print("=" * 60) print("3. 套餐构成（covered_codes）与枚举深度（max_sides）") print("=" * 60) combo = Item( code="COMBO1", name="某三件套", price=39.0, parts=[ Item(code="M001", name="麦辣鸡腿堡", price=0.0), Item(code="S001", name="中薯条", price=0.0), Item(code="S002", name="可口可乐(中)", price=0.0), ], ) assert combo.covered_codes() == {"COMBO1", "M001", "S001", "S002"}, \ "套餐应覆盖「自身编码 + 内含单品编码」" plan = Plan(items=[combo]) assert {"M001", "S001", "S002"} <= plan.covered_codes(), \ "Plan.covered_codes() 应把套餐内含单品算进来" # 内含单品只是元数据，不参与计价 assert plan.goods_amount() == 39.0, "套餐内含单品不应影响套餐计价" print(f" 套餐覆盖编码：{sorted(combo.covered_codes())}") print(" [PASS] 套餐知道自己在装什么，且不影响计价") one_main = [Item(code="M1", name="主食", price=10.0)] two_sides = [ Item(code="S1", name="配餐A", price=3.0, is_side=True), Item(code="S2", name="配餐B", price=4.0, is_side=True), ] shallow = build_base_plans(one_main, two_sides, [], max_sides=1) deep = build_base_plans(one_main, two_sides, [], max_sides=2) assert len(shallow) == 3, f"max_sides=1 应产出「主食」+2 个两件套 = 3，实际 {len(shallow)}" assert len(deep) == 4, f"max_sides=2 应多出「主食+两个配餐」，实际 {len(deep)}" print(f" max_sides=1 → {len(shallow)} 个方案；max_sides=2 → {len(deep)} 个") print(" [PASS] 需求要三样齐全时，必须 max_sides≥2 才枚举得出对照组合") print() def check_budget(mains, sides) -> None: """校验预算语义：区分「预算内没有」与「有但更贵」。""" print("=" * 60) print("4. 预算：区分「预算内没有」与「有方案但更贵」") print("=" * 60) def wants_burger_and_fries(plan: Plan) -> bool: """需求：麦辣鸡腿堡 + 中薯条，两样齐全。""" codes = plan.covered_codes() return "M001" in codes and "S001" in codes # 满足需求的最便宜组合是 22 + 12 = 34 元，预算给 30 元 tight = recommend( mains=mains, sides=sides, combos=[], coupons=[], price_fn=mock_price_fn, budget=30.0, demand=wants_burger_and_fries, ) assert tight["recommended"] == [], "预算内没有方案时 recommended 应为空" assert tight["over_budget"], "超预算的对照方案应单独返回" assert tight["best"] is None, "预算内没有方案时不应给出 best" print(f" 预算 30：预算内 0 个；超预算对照 {len(tight['over_budget'])} 个" f"（最便宜 {tight['over_budget'][0].payable} 元）") render_txt = render(tight) assert "预算内没有可用方案" in render_txt, "应如实说明预算内没有方案" print(" [PASS] 渲染如实说明「预算内没有」，而不是拿超预算方案冒充推荐") # within_budget 不再兜底把超预算方案当成推荐 single = [mock_price_fn(Plan(items=[mains[0]]))] assert within_budget(single, 1.0) == [], \ "全部超预算时应返回空列表，而不是兜底返回最便宜的" print(" [PASS] within_budget 不再兜底（超预算 ≠ 推荐）") print() def main() -> None: """跑一遍省钱枚举的完整场景并断言结果。""" mains = [ Item(code="M001", name="麦辣鸡腿堡", price=22.0, category="汉堡"), Item(code="M002", name="双层吉士汉堡", price=25.0, category="汉堡"), ] sides = [ Item(code="S001", name="中薯条", price=12.0, is_side=True), Item(code="S002", name="可口可乐(中)", price=11.0, is_side=True), Item(code="S003", name="香芋派", price=8.0, is_side=True), ] combos = [ [Item(code="M001", name="麦辣鸡腿堡", price=22.0), Item(code="S001", name="中薯条", price=12.0), Item(code="S002", name="可口可乐(中)", price=11.0)], ] coupons = [ Coupon(id="C1", name="满20减3", threshold=20.0, discount=3.0), Coupon(id="C2", name="满50减12", threshold=50.0, discount=12.0), ] print("=" * 60) print("1. 基础推荐（含预算筛选与凑单陷阱）") print("=" * 60) result = recommend( mains=mains, sides=sides, combos=combos, coupons=coupons, price_fn=mock_price_fn, budget=45.0, top_k=3, ) print(render(result)) # --- 断言：逻辑正确性 --- assert result["best"] is not None, "应当至少产出一个推荐方案" rec = result["recommended"] assert all(p.payable <= 45.0 for p in rec), "预算筛选失效：存在超预算方案" assert rec == sorted(rec, key=lambda p: p.payable), "推荐结果未按实付升序排列" assert all(p.payable > 45.0 for p in result["over_budget"]), \ "over_budget 里混进了预算内的方案" warnings = result["warnings"] assert len(warnings) > 0, "应当检测出凑单陷阱（满50减12）" # 陷阱方案的商品数必须多于替代方案，且实付不低于替代方案 for w in warnings: assert w["extra_items"] > 0, "陷阱判定错误：多买的商品数应为正" assert w["trap_payable"] >= w["alternative_payable"], "陷阱判定错误：陷阱方案应不更便宜" print(f"\n[PASS] 推荐方案 {len(rec)} 个，全部在预算内且已升序排列") print(f"[PASS] 超预算对照 {len(result['over_budget'])} 个，已与推荐分开") print(f"[PASS] 检测到 {len(warnings)} 个凑单陷阱") print(f"[PASS] 最优方案实付：{result['best'].payable} 元 —— {result['best'].label}") print() check_demand(mains, sides, combos, coupons) check_combo_parts() check_budget(mains, sides) print("=" * 60) print("全部省钱逻辑测试通过") print("=" * 60) if __name__ == "__main__": main()