"""知识块粒度选择的单元测试。

本文件当前只覆盖**检索层**（`KnowledgeSearchService`）的粒度归并。原先另有 4 条用例覆盖
`CustomerServiceAgent._prefer_section`（「整节块」与「行级子块」之间的取舍），已随客服
Agent 模块一并移除——重建客服 Agent 时须把该判据连同用例一起带回来。

检索层用例的背景是实测的三次翻车，每条判据都对应其中一次：

1. 客户问「起投多少」和「风险高吗」时命中同一块（整个产品小节），拿到**完全相同**的
   整节内容，看起来像客服没听懂问题——所以把表格行拆成了行级子块。
2. 拆细之后「介绍一下」又被某一行抢答（返回"产品期限 90天封闭期"）——所以要能换回整节。
3. 两次判据写错：用"含连字符"认子块时，整节块自己的编号 PROD-901 被误判成子块；
   用"不含两位数字后缀"认整节块时，FAQ 块全被误判成整节块、把正确答案挤出了 top1。

第 4 次翻车（2026-09-15）与"同节兄弟子块互相打平"有关：`doc_id` 去重挡不住
`POL-AST-009-07` 与 `POL-AST-009-12` 这种**同父不同子**，实测它们把「风险评估问卷怎么评分」
的 top1/次优差压到 0.002 → 客服判并列转人工。
"""

from app.service.knowledge_search_service import KnowledgeHit, KnowledgeSearchService


def hit(doc_id: str, score: float, content: str = "正文") -> KnowledgeHit:
    return KnowledgeHit(doc_id=doc_id, title=doc_id, content=content, score=score)


def test_parent_of_recognises_row_blocks() -> None:
    assert KnowledgeSearchService._parent_of("PROD-007-04") == "PROD-007"


def test_parent_of_rejects_section_blocks() -> None:
    """整节块的编号本身就含连字符（PROD-901），不能被当成子块。"""
    assert KnowledgeSearchService._parent_of("PROD-901") is None
    assert KnowledgeSearchService._parent_of("FAQ-0016") is None
    assert KnowledgeSearchService._parent_of("HNW-003") is None


# --- 同节兄弟子块归并（2026-09-15） -------------------------------------------


def test_sibling_subblocks_of_one_section_collapse_to_the_highest_scoring_one() -> None:
    """同一节的多个子块是"同一答案的不同细节"，不是并列候选：只留最高分那条。"""
    hits = [
        hit("POL-AST-009-12", 0.7359),
        hit("POL-AST-009-07", 0.7346),
        hit("POL-AST-009-19", 0.7340),
        hit("POL-AST-009-51", 0.7340),
    ]

    merged = KnowledgeSearchService._merge_sibling_subblocks(hits)

    assert [item.doc_id for item in merged] == ["POL-AST-009-12"]


def test_merge_keeps_one_block_per_section() -> None:
    """**不同**父块各自的最高分子块都要留下：它们是真正不同的候选。"""
    hits = [
        hit("PROD-007-04", 0.86),
        hit("PROD-007-05", 0.85),
        hit("HNW-005-02", 0.80),
        hit("HNW-005-01", 0.79),
    ]

    merged = KnowledgeSearchService._merge_sibling_subblocks(hits)

    assert [item.doc_id for item in merged] == ["PROD-007-04", "HNW-005-02"]


def test_merge_leaves_plain_blocks_alone() -> None:
    """FAQ / 政策 / 公司信息这类块本身就是细粒度答案：它们之间打平是真的多个候选，
    **不能**合并（否则"存在并列"这个信号会被抹掉，客服会硬答一个巧合高分）。"""
    hits = [
        hit("FAQ-0016", 0.85),
        hit("FAQ-0015", 0.84),
        hit("POL-SPM-010", 0.83),
        hit("POL-AST-009", 0.82),
    ]

    assert KnowledgeSearchService._merge_sibling_subblocks(hits) == hits


def test_merge_preserves_score_order_and_the_parent_block() -> None:
    """归并不改顺序；父块（整节）不受影响，仍会按保底名额回到候选里。"""
    hits = [
        hit("PROD-007-04", 0.86),
        hit("PROD-007", 0.77),          # 父块：整节，`_parent_of` 为 None
        hit("PROD-007-05", 0.75),
        hit("FAQ-0015", 0.60),
    ]

    merged = KnowledgeSearchService._merge_sibling_subblocks(hits)

    assert [item.doc_id for item in merged] == ["PROD-007-04", "PROD-007", "FAQ-0015"]
    assert [item.score for item in merged] == [0.86, 0.77, 0.60]


def test_merge_is_idempotent() -> None:
    """重复调用不得继续删东西（幂等，便于以后在别处复用）。"""
    hits = [hit("PROD-007-04", 0.86), hit("PROD-007-05", 0.75), hit("FAQ-0015", 0.60)]

    once = KnowledgeSearchService._merge_sibling_subblocks(hits)
    twice = KnowledgeSearchService._merge_sibling_subblocks(once)

    assert twice == once


def test_merge_of_empty_list_is_empty() -> None:
    assert KnowledgeSearchService._merge_sibling_subblocks([]) == []