a2bd3960e1
- report_meta：回款天数区间桶（含可空/负值兜底桶）、客户类型标签、导出上限 - schemas/report：明细/汇总查询入参与分页响应，金额比例全程 Decimal - report_repository：报表专用只读聚合，显式排除台账聚合行、口径与月结区分 - constants：新增报表统计模块与导出动作常量 Co-Authored-By: Claude Code <noreply@anthropic.com>
77 lines
3.1 KiB
Python
77 lines
3.1 KiB
Python
"""报表统计元数据（区间桶、中文标签、导出上限）——叶子模块。
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刻意不导入 app 内任何模块：report_service 与 report_export 都要消费这里的
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定义，若它反向依赖 service 层会形成循环导入；也不走参数体系——FR-031
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「区间可配置」一期以代码常量落地（参数化需注册表 + 迁移 + init-dev +
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registry 测试四处同改，业务未提诉求前不做），配置缝收在
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get_payment_day_buckets 单函数，未来在线可配时只改该函数。
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"""
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from dataclasses import dataclass
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@dataclass(frozen=True, slots=True)
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class DayBucket:
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"""回款天数区间档位：min/max 均为 None 表示「无回款天数」兜底桶。"""
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code: str
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label: str
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min_days: int | None
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max_days: int | None
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# 8 档区间（催办清单 R2 默认口径）。两个兜底桶必须保留：payment_days 可空
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# （回款未指定发货批次）且可为负（先款后货照实负值），缺了它们这些行会
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# 静默消失、金额对不上。
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PAYMENT_DAY_BUCKETS: tuple[DayBucket, ...] = (
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DayBucket("le0", "≤0 天（先款后货）", None, 0),
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DayBucket("1-30", "1-30 天", 1, 30),
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DayBucket("31-60", "31-60 天", 31, 60),
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DayBucket("61-90", "61-90 天", 61, 90),
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DayBucket("91-180", "91-180 天", 91, 180),
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DayBucket("181-360", "181-360 天", 181, 360),
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DayBucket("361+", "361 天以上", 361, None),
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DayBucket("none", "无回款天数", None, None),
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)
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CUSTOMER_TYPE_LABELS: dict[str, str] = {
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"personal": "个人户",
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"manager_personal": "经理个人户",
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"enterprise": "企业户",
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}
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# mixed 不在 core.constants.PRODUCT_CATEGORIES（那只是可维护的档案类别），
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# 但台账行可以是 mixed（混合业务行级明细入 snapshot_json），报表必须能标它
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PRODUCT_CATEGORY_LABELS: dict[str, str] = {
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"municipal_water": "市政水利",
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"rail_transit": "轨道交通",
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"mixed": "混合业务",
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}
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REPORT_TEMPLATE_VERSION = "V1.0"
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# 导出超限即拒绝（需求「大数据量分批/超限提示」二选一取后者）：普通
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# Workbook 全量载入内存，2 万行约数十 MB，超过该量级必须引导按月导出
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REPORT_EXPORT_MAX_ROWS = 20000
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def get_payment_day_buckets() -> tuple[DayBucket, ...]:
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"""区间档位唯一出口：前端下拉与分桶都以本函数返回为准，禁止硬编码。
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未来「区间在线可配」时改此处读参数即可，调用方与前端零改动。
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"""
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return PAYMENT_DAY_BUCKETS
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def bucket_of(payment_days: int | None) -> DayBucket:
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"""把单笔回款天数落入对应区间桶；未匹配任何区间按「无回款天数」兜底。"""
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if payment_days is None:
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return next(b for b in PAYMENT_DAY_BUCKETS if b.code == "none")
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for bucket in PAYMENT_DAY_BUCKETS:
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if bucket.code == "none":
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continue
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lower_ok = bucket.min_days is None or payment_days >= bucket.min_days
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upper_ok = bucket.max_days is None or payment_days <= bucket.max_days
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if lower_ok and upper_ok:
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return bucket
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return next(b for b in PAYMENT_DAY_BUCKETS if b.code == "none")
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