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选择的列不同时，差异说明整段退回「两条 SQL 的写法不同，未能归纳成具体差异」（真实验收 E2 的兜底比例 45%）：只差在选择 列表上时一条具体说法都没有；多选的列里有 sqlglot 不认识的函数（SQLite 的 printf 这类）时整段不算简单形状，WHERE 里 真实的差别也不写。 - ledger/match：按引用的列投影过、仍没证明的候选，差在哪用投影之后那次比较的结果；答案多选了列的，在最前面写一句 「这次多查了 N 列，不影响这个数」（aspect=projection），然后照常说条件、时间、聚合上的不同。E1 的差异类型不数这一句。 - sql_semantics 的 _Explainer._left_on_times：两边来源一样、只差在一个 LEFT JOIN 的 ON 里的时间条件时，说「这次用 LEFT JOIN 关联 orders 表时只关联 … 的行，口径没有这个条件」「…时对 paid_at 的条件是 …，口径是 …」「…时按 paid_at 筛时间， 口径按 created_at」。说得准才说：差的每个条件都是只涉及可空一侧那张表的时间条件、那张表只出现在结果里（WHERE、 HAVING、别的连接没用到，SELECT 列表用到了）、同一列的按取值算过确实不同、单边的不是恒为真；套着日期函数认不准的、 反连接、结果里没用到可空一侧的、不是时间条件的照实兜底。aspect 写 join（说的是关联条件），双盲复算不会把它当业务日期 写进数据目录。只改说明，判等价不变。 测试：tests/test_ledger_projection.py（没证明时先说多查了几列、再说少了 status 条件，不再兜底）、 tests/test_sql_left_on_times.py（四种说得清的、在虚构库上结果确实不同；四种说不清的照实兜底；判等价的结论不变）。 去掉这两处改动再跑，各自的用例失败（2 条、8 条）。
83 lines
4.7 KiB
Python
83 lines
4.7 KiB
Python
"""日期条件写在 LEFT JOIN 的 ON 里时的差异说明（data/sql_semantics.py 的 _Explainer._left_on_times）。
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以前两边来源一样、只差在一个 LEFT JOIN 的 ON 里的日期条件时，整段退回「两条 SQL 的写法不同，未能归纳成具体差异」。现在
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说得准的照实说（aspect 是 join：说的是关联条件，不会被当成业务日期写进数据目录），说不准的照实兜底。只改说明：判等价的
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结论一个不变；说出来的不同在虚构库上确实让结果不一样。
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"""
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from __future__ import annotations
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import pytest
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from app.data.sql_semantics import explain_sql_diff
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from tests.fixtures import sql_pairs
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SCHEMA = sql_pairs.schema_cache()
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BASE = "SELECT u.city, SUM(o.amount) AS s FROM users u LEFT JOIN orders o ON o.user_id = u.id{on} GROUP BY u.city"
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SAID = [
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# 一边多一个日期条件
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(BASE.format(on=" AND o.paid_at >= '2026-09-10'"), BASE.format(on=""),
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"这次用 LEFT JOIN 关联 orders 表时只关联 orders.paid_at >= '2026-09-10' 的行，口径没有这个条件"),
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(BASE.format(on=""), BASE.format(on=" AND o.created_at < '2026-09-20'"),
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"口径用 LEFT JOIN 关联 orders 表时只关联 orders.created_at < '2026-09-20' 的行，这次没有这个条件"),
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# 按不同的时间列筛
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(BASE.format(on=" AND o.paid_at >= '2026-09-10'"), BASE.format(on=" AND o.created_at >= '2026-09-10'"),
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"这次用 LEFT JOIN 关联 orders 表时按 paid_at 筛时间，口径按 created_at"),
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# 同一列、范围不同
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(BASE.format(on=" AND o.paid_at >= '2026-09-10' AND o.paid_at < '2026-10-01'"),
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BASE.format(on=" AND o.paid_at >= '2026-09-01' AND o.paid_at < '2026-10-01'"),
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"这次用 LEFT JOIN 关联 orders 表时对 paid_at 的条件是 orders.paid_at >= '2026-09-10' AND orders.paid_at < "
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"'2026-10-01'，口径是 orders.paid_at >= '2026-09-01' AND orders.paid_at < '2026-10-01'"),
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]
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@pytest.fixture(scope="module")
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def databases():
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conns = sql_pairs.build_databases()
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yield conns
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for c in conns:
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c.close()
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@pytest.mark.parametrize("a, b, text", SAID, ids=["extra-on-ours", "extra-on-caliber", "other-column", "range"])
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@pytest.mark.parametrize("schema", [None, SCHEMA], ids=["bare", "schema"])
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def test_date_conditions_in_a_left_join_on_are_said(a, b, text, schema, databases):
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diff = explain_sql_diff(a, b, "sqlite", schema=schema, labels=("这次", "口径"))
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assert diff.equivalent is False and diff.fallback is False
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assert [(i.aspect, i.text) for i in diff.items] == [("join", text)]
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# 说出来的不同是真的：在虚构库的某组数据上两条结果确实不一样
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assert any(sql_pairs.run(c, a) != sql_pairs.run(c, b) for c in databases)
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@pytest.mark.parametrize("a, b", [
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# 套着日期函数，按取值认不准两边是不是一样：照实兜底
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(BASE.format(on=" AND DATE(o.paid_at) >= '2026-09-10' AND DATE(o.paid_at) <= '2026-09-30'"),
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BASE.format(on=" AND o.paid_at >= '2026-09-10' AND o.paid_at < '2026-10-01'")),
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# 可空一侧在 WHERE 里用到了（反连接）：说的就不是时间窗口了
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("SELECT u.city, COUNT(*) AS n FROM users u LEFT JOIN orders o ON o.user_id = u.id AND o.paid_at >= '2026-09-10' "
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"WHERE o.id IS NULL GROUP BY u.city",
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"SELECT u.city, COUNT(*) AS n FROM users u LEFT JOIN orders o ON o.user_id = u.id WHERE o.id IS NULL "
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"GROUP BY u.city"),
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# 结果里没用到可空一侧的列：ON 里的不同只影响行会不会重复，说不准
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("SELECT u.city, COUNT(*) AS n FROM users u LEFT JOIN orders o ON o.user_id = u.id AND o.paid_at >= '2026-09-10' "
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"GROUP BY u.city",
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"SELECT u.city, COUNT(*) AS n FROM users u LEFT JOIN orders o ON o.user_id = u.id GROUP BY u.city"),
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# ON 里多出来的不是时间条件
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(BASE.format(on=" AND o.status = 'paid'"), BASE.format(on="")),
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], ids=["date-function", "anti-join", "not-in-output", "not-time"])
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def test_what_cannot_be_said_precisely_still_gets_the_fallback(a, b):
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diff = explain_sql_diff(a, b, "sqlite", labels=("这次", "口径"))
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assert diff.equivalent is False and diff.fallback is True
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assert not [i for i in diff.items if i.aspect == "join"]
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@pytest.mark.parametrize("a, b, same", [
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(BASE.format(on=" AND o.paid_at >= '2026-09-10'"), BASE.format(on=" AND o.paid_at >= '2026-09-10'"), True),
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("SELECT u.city, SUM(o.amount) AS s FROM users u LEFT JOIN orders o ON o.paid_at >= '2026-09-10' AND "
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"o.user_id = u.id GROUP BY u.city", BASE.format(on=" AND o.paid_at >= '2026-09-10'"), True),
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*[(a, b, False) for a, b, _ in SAID],
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])
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def test_only_the_explanation_changed_not_the_verdict(a, b, same):
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for schema in (None, SCHEMA):
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assert explain_sql_diff(a, b, "sqlite", schema=schema).equivalent is same
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