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SQL 语义复核修正（第 4 轮）：外连接按主表比、算出来的列按定义比、时间列透过 CTE 认、筛的先后照报、没用到的 CTE 不算，同名却不是同一个的派生来源说清各是什么
上一轮（db91c9e）给派生来源加了结构标识、按结构配对之后，复验还有一批用例返回空列表或者报得自相矛盾。按根因分几类改：
外连接：joins 第二遍按涉及的来源配对、不分方向，orders LEFT JOIN refunds 和 refunds LEFT JOIN orders 配成一对后类型、条件都一样，一句不说（db91c9e 引入的回归），RIGHT JOIN 对调、两个 CTE 对调、子查询写法同样漏。现在 JoinFacet 记下主表（keep）和可空那一侧（nullable），LEFT / RIGHT 按主表比：对调了报「原流程以 orders 为主表外连接 refunds，盲算以 refunds 为主表外连接 orders」，LEFT JOIN 和两边对调的 RIGHT JOIN 是同一回事、不再报成类型不同（原来列为做不到）。同一段 WHERE 里拿可空那一侧的列比较、IN、LIKE、BETWEEN、IS NOT NULL 的（JoinFacet.rejects），这次外连接按内连接比：盲算把可空一侧的条件写进 WHERE、LEFT JOIN 实际成了内连接的照报；IS NULL、OR、套在 COALESCE / CASE 里的不认。
算出来的列（Computed，新的 expression 一面）：聚合只按去重后的 (函数, 参数) 集合比，SUM(p.total) - SUM(r.total) 对反过来、退款率分子分母对调报不出；派生来源里不是聚合的列不进任何方面，net.v 谁减谁、x.total 取的是付款还是退款合计、amount * 0.9 对 0.8、CASE 的口径、窗口函数的分区和排序、RANK 对 ROW_NUMBER 都是空列表。现在最外层选出来的列（先剥掉 ROUND、CAST、COALESCE 给默认值这类外壳）按算法比，用到的聚合一样、位置换了照报；派生来源里被外面用到、又不是原样透出一列的列按定义比。只差在聚合上、聚合那一面已经报了的，按透过去的那一列筛时间、时间那一面已经报了的，不再报第二遍。比较时加法、乘法的两项排好先后，DATE()、CAST(… AS DATE)、::date 统一成一种写法，免得写法不同被报成算法不同。
时间：时间条件的列透过「同一列套着日期函数、CAST、MIN / MAX」算出来的派生列，一直换到实表（TimePredicate.via 记下路过的列）：日期表达式写在 CTE 里，按支付时间还是下单时间筛照样报出来，带上表名、列名给调用方对照数据目录（三个方言的 DATE()、CAST … BETWEEN、两个同名 CTE 对调后的 MAX(paid_at) 对 MAX(refunded_at)）。
结构标识和配对：_Derived 里往里套着的派生来源原来一律写「@」，net 是 paid.total - refunded.total 还是反过来写出来一样。现在按结构写的 shape 写法里引用派生来源的地方写它自己的结构标识（里面那段按 shape 写出来再取摘要），配对之前比较的两边也按结构标识排先后；配对之后仍按共同编号排（配成一对的两个派生来源里面稍有不同时结构标识就不同，按它排会把同一个比较排成两个方向）。_align 同一级先看条件重合、再看名字：两边都叫 paid、一边筛已付一边筛未付时不再按名字配反。配成一对的派生来源里两边起名不同的列按定义配成同一列（total 对 sum_amt，原来列为做不到；_renamed），两边各剩一列时也配上，定义的不同照报。
筛的先后：派生来源取前几行（LIMIT、OFFSET）或算了窗口函数（这一列又不是分区键，含 DISTINCT ON）时，外面的条件不再透成里面那张表的条件；Part 另记一路透到底的写法（pierce）和挡住它的是什么（stage），两边只差在这里时报「原流程先算 t 的窗口函数、再按 status = 'paid' 筛选，盲算先按 status = 'paid' 筛选、再算 t 的窗口函数」「先取 t 的前 10 行、再筛」，时间条件同样。
没用到的 CTE：原来逐段收集连定义了没用的 CTE 里的条件也收，盲算定义了只筛已付的 CTE 却直接查 orders 报不出来，没用的 CTE 里多一个条件又会被报成多筛了一遍。现在只收从最外层一路引用到的段（_Statement._enliven）；例外是两边都定义了同一个 CTE、一边把它关联进来一边没用到，没用到的那边照样收它，免得它里面的聚合、分组被说成一边独有（连接那一面已经说了关联了它）。
说清楚：两边同名的派生来源其实不是同一个时（t1 / t2、a / b 定义对调），「按 t1.total > 100 筛选」两边写法一模一样，读起来自相矛盾。Part 记下用到的派生来源各列的来历（origins，一直换到实表的列），给人看的写法依次退到一律写表名、带上来历（t1.total 即 SUM(orders.amount)）、来历也一样（两个 CTE 只差里面的条件）时说「盲算的 paid 对应原流程的 open」。
自己关联自己：同一个 CTE、同一张表在一段里出现两次，两个别名按各自在关联条件里出的列区分（_Source.role），b1.total > b2.total 写反了报得出来，换了别名、换了关联先后不报（原来列为做不到）。CTE、子查询自己写了列名的（WITH t(n) AS …，递归 CTE 常这么写）按写的列名认列。
模块说明、Part / JoinFacet / TimePredicate / Computed / Facets / FacetDiff、unmatched、joins 的文档按现在的做法改写，做不到的照实列出：两边起名不同、各多出不止一列且定义都不同的列仍算两列；全换了的派生来源不配对；条件改得一个都不重合又各起了名字的同形派生来源只能按先后配；自连接两个别名在关联条件里出的列一样时分不出；三项以上的加法乘法换序、括号展开算不同算法；QUALIFY、GROUP BY 之后才筛的先后不认；外连接链里连带变成内连接的、套在函数里或写在外层 WHERE 的不认；不是 facets 算出来的 Facets 只能按位置编号比。
测试：复验清单里的用例逐个补成测试（各方言、CTE 和子查询写法），另加意思一样、不该报的反例（LEFT / RIGHT 对调、WHERE 筛主表、外连接链换序、0.9 * amount、DATE 对 CAST、按分区键在窗口前后筛、两边都先取前 10 行、列起名不同、自连接换别名换先后、没用到的 CTE）。新加 64 项，其中 48 项在 db91c9e 上失败，其余是防误报的，两边都通过；原有 128 项和 test_verify_agent 照旧通过，全量 7113 项通过。复验清单最后一条只到「WITH base AS (SE」就截断了，看不到全文，按开头和已知缺口一并处理了 CTE 自连接。
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@@ -9,7 +9,10 @@
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- diff_facets：每处不同一句中文：「A 按支付时间 paid_at 筛时间，B 按下单时间 created_at」「A 排除了
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- diff_facets：每处不同一句中文：「A 按支付时间 paid_at 筛时间，B 按下单时间 created_at」「A 排除了
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is_deleted = 1 的行，B 没有」「A 对 order_id 去重计数，B 没去重」……写法不同、口径相同的不报。两边的派生来源先按
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is_deleted = 1 的行，B 没有」「A 对 order_id 去重计数，B 没去重」……写法不同、口径相同的不报。两边的派生来源先按
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结构（读的表、各列的定义、名字）配好对再比：一边多出一个排在前面的 CTE、内连接换了先后不报，两个 CTE 算出同名
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结构（读的表、各列的定义、名字）配好对再比：一边多出一个排在前面的 CTE、内连接换了先后不报，两个 CTE 算出同名
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的列时外层用的是哪一个照报
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的列时外层用的是哪一个照报；外连接按主表比（LEFT / RIGHT 对调不报，主表换了、被 WHERE 变成内连接照报）；算出来
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的列按算法比（最外层的算式、窗口函数，CTE 里被外面用到的列的定义）；时间列透过 CTE 里的日期表达式认；取前几行、
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窗口函数前后筛的不同照报；没用到的 CTE 不算；两边同名却不是同一个的派生来源说清各是什么；自己关联自己的两个
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别名分得开
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夹具全是虚构的表：订单 orders、退款 refunds、门店 stores、用户 users、付款 payments，以及一张中文表「订单」。
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夹具全是虚构的表：订单 orders、退款 refunds、门店 stores、用户 users、付款 payments，以及一张中文表「订单」。
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"""
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"""
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@@ -885,6 +888,453 @@ def test_which_rows_are_taken_is_compared(a, b, text):
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assert texts("SELECT SUM(amount) FROM orders ORDER BY 1 DESC", "SELECT SUM(amount) FROM orders") == []
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assert texts("SELECT SUM(amount) FROM orders ORDER BY 1 DESC", "SELECT SUM(amount) FROM orders") == []
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# --------------------------------------------------------------------------
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# diff_facets：外连接以哪一侧为主表
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# --------------------------------------------------------------------------
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LABELS = ("原流程", "盲算")
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@pytest.mark.parametrize("dialect,a,b,expected", [
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("sqlite", "SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id",
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"SELECT COUNT(*) FROM refunds r LEFT JOIN orders o ON o.id = r.order_id",
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["原流程以 orders 为主表外连接 refunds，盲算以 refunds 为主表外连接 orders"]),
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# 「没有退款的订单数」对「order_id 为空的退款数」：IS NULL 不会把外连接变成内连接，差的是主表
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("sqlite",
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"SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id WHERE r.order_id IS NULL",
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"SELECT COUNT(*) FROM refunds r LEFT JOIN orders o ON o.id = r.order_id WHERE r.order_id IS NULL",
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["原流程以 orders 为主表外连接 refunds，盲算以 refunds 为主表外连接 orders"]),
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("postgres", "SELECT COUNT(*) FROM orders o RIGHT JOIN refunds r ON r.order_id = o.id",
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"SELECT COUNT(*) FROM refunds r RIGHT JOIN orders o ON o.id = r.order_id",
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["原流程以 refunds 为主表外连接 orders，盲算以 orders 为主表外连接 refunds"]),
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("postgres", PAID_REFUNDED + "SELECT COUNT(*) FROM paid p LEFT JOIN refunded r ON r.user_id = p.user_id",
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PAID_REFUNDED + "SELECT COUNT(*) FROM refunded r LEFT JOIN paid p ON p.user_id = r.user_id",
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["原流程以 paid 为主表外连接 refunded，盲算以 refunded 为主表外连接 paid"]),
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("mysql",
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"SELECT COUNT(*) FROM (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) p "
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"LEFT JOIN (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id) r ON r.user_id = p.user_id",
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"SELECT COUNT(*) FROM (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id) r "
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"LEFT JOIN (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) p ON p.user_id = r.user_id",
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["原流程以 p 为主表外连接 r，盲算以 r 为主表外连接 p"]),
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], ids=["left", "left-anti", "right", "cte", "subquery"])
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def test_which_side_an_outer_join_keeps_is_compared(dialect, a, b, expected):
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"""外连接两边对调，保留的就是另一张表的全部行（有没有退款的订单都算，还是有没有订单的退款都算）：按主表比，
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关联的表、条件一样也要报出来。"""
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assert texts(a, b, dialect, labels=LABELS) == expected
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@pytest.mark.parametrize("dialect,a,b", [
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("sqlite", "SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id",
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"SELECT COUNT(*) FROM refunds r RIGHT JOIN orders o ON o.id = r.order_id"),
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("postgres", PAID_REFUNDED + "SELECT COUNT(*) FROM paid p LEFT JOIN refunded r ON r.user_id = p.user_id",
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PAID_REFUNDED + "SELECT COUNT(*) FROM refunded r RIGHT JOIN paid p ON p.user_id = r.user_id"),
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# 外连接后又在 WHERE 里筛可空那一侧：等于内连接加同一个条件
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("sqlite", "SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id WHERE r.state = 'done'",
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"SELECT COUNT(*) FROM orders o JOIN refunds r ON r.order_id = o.id WHERE r.state = 'done'"),
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# 筛的是主表那一侧，外连接照旧
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("sqlite", "SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id WHERE o.status = 'paid'",
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"SELECT COUNT(*) FROM orders AS o LEFT OUTER JOIN refunds AS r ON o.id = r.order_id WHERE o.status = 'paid'"),
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# 外连接链换了先后，各自的主表不变
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("sqlite",
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"SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id LEFT JOIN users u ON u.id = o.user_id",
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"SELECT COUNT(*) FROM orders o LEFT JOIN users u ON u.id = o.user_id LEFT JOIN refunds r ON r.order_id = o.id"),
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], ids=["left-vs-right", "left-vs-right-cte", "rejected-left-vs-inner", "filter-on-kept-side", "chain-reordered"])
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def test_the_same_outer_join_written_differently_is_no_difference(dialect, a, b):
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"""orders LEFT JOIN refunds 和 refunds RIGHT JOIN orders 是同一次连接：按主表比，不按写的是 LEFT 还是 RIGHT。"""
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assert texts(a, b, dialect) == []
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@pytest.mark.parametrize("dialect,a,b,expected", [
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("sqlite", "SELECT COUNT(*) FROM orders o LEFT JOIN (SELECT * FROM refunds WHERE state = 'done') r "
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"ON r.order_id = o.id",
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"SELECT COUNT(*) FROM orders o LEFT JOIN (SELECT * FROM refunds) r ON r.order_id = o.id WHERE r.state = 'done'",
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["原流程以 orders 为主表外连接 r，盲算的 WHERE 里有 refunds.state = 'done'，把关联不上 r 的行筛掉了，"
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"LEFT JOIN 实际成了 INNER JOIN"]),
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("mysql", "WITH done AS (SELECT * FROM refunds WHERE state = 'done') "
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"SELECT SUM(o.amount) FROM orders o LEFT JOIN done d ON d.order_id = o.id",
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"WITH done AS (SELECT * FROM refunds) "
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"SELECT SUM(o.amount) FROM orders o LEFT JOIN done d ON d.order_id = o.id WHERE d.state = 'done'",
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["原流程以 orders 为主表外连接 done，盲算的 WHERE 里有 refunds.state = 'done'，把关联不上 done 的行筛掉了，"
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"LEFT JOIN 实际成了 INNER JOIN"]),
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# 盲算以 refunds 为主表，WHERE 又筛掉了没对上订单的行：没有退款的已付订单原流程算、盲算不算
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("postgres",
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"SELECT SUM(o.amount) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id WHERE o.status = 'paid'",
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"SELECT SUM(o.amount) FROM refunds r LEFT JOIN orders o ON o.id = r.order_id WHERE o.status = 'paid'",
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["原流程以 orders 为主表外连接 refunds，盲算的 WHERE 里有 orders.status = 'paid'，把关联不上 orders 的行筛掉了，"
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"LEFT JOIN 实际成了 INNER JOIN"]),
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("postgres", "SELECT COUNT(*) FROM orders o RIGHT JOIN refunds r ON r.order_id = o.id",
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"SELECT COUNT(*) FROM orders o RIGHT JOIN refunds r ON r.order_id = o.id WHERE o.status = 'paid'",
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["盲算只保留 status = 'paid' 的行，原流程没有这个条件",
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"原流程以 refunds 为主表外连接 orders，盲算的 WHERE 里有 orders.status = 'paid'，把关联不上 orders 的行筛掉了，"
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"RIGHT JOIN 实际成了 INNER JOIN"]),
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], ids=["subquery", "cte", "swapped-and-filtered", "right"])
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def test_an_outer_join_turned_inner_by_where_is_reported(dialect, a, b, expected):
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"""可空那一侧的条件写在 ON / 子查询里，和写在 WHERE 里不是一回事：WHERE 里拿空值比不成立，没关联上的行被筛掉，
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LEFT JOIN 实际成了内连接。条件本身两边一样（透过子查询都是 refunds.state = 'done'），只报连接。"""
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assert texts(a, b, dialect, labels=LABELS) == expected
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def test_only_null_rejecting_where_conditions_turn_an_outer_join_inner():
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"""IS NULL、和 IS NULL 用 OR 连着的、套在 COALESCE 里的，没关联上的行照样留着：外连接不变（宁可漏报）。"""
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left = "SELECT COUNT(*) FROM orders o LEFT JOIN refunds r ON r.order_id = o.id"
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inner = "SELECT COUNT(*) FROM orders o JOIN refunds r ON r.order_id = o.id"
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either = " WHERE r.state = 'done' OR r.id IS NULL"
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assert texts(left + either, inner + either) == ["A 用 LEFT JOIN 关联 refunds 表，B 用 INNER JOIN"]
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assert aspects(left + " WHERE COALESCE(r.amount, 0) < 100", left) == ["filter"]
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# --------------------------------------------------------------------------
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# diff_facets：算出来的列
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# --------------------------------------------------------------------------
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@pytest.mark.parametrize("dialect,a,b,expected", [
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("sqlite", PAID_REFUNDED + f"SELECT SUM(p.total) - SUM(r.total) {PR_JOIN}",
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PAID_REFUNDED + f"SELECT SUM(r.total) - SUM(p.total) {PR_JOIN}",
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["原流程算的是 SUM(paid.total) - SUM(refunded.total)，盲算算的是 SUM(refunded.total) - SUM(paid.total)"]),
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# 退款率的分子分母对调
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("postgres", f"SELECT SUM(r.total) / SUM(p.total) {PR_SUBQUERIES}",
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f"SELECT SUM(p.total) / SUM(r.total) {PR_SUBQUERIES}",
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["原流程算的是 SUM(r.total) / SUM(p.total)，盲算算的是 SUM(p.total) / SUM(r.total)"]),
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# 窗口：每个用户的合计还是每家门店的合计
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("sqlite", "SELECT order_id, SUM(amount) OVER (PARTITION BY user_id) AS s FROM orders",
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"SELECT order_id, SUM(amount) OVER (PARTITION BY store_id) AS s FROM orders",
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["原流程算的是 SUM(amount) OVER (PARTITION BY user_id)，盲算算的是 SUM(amount) OVER (PARTITION BY store_id)"]),
|
||||||
|
# 同一个合计再算一下（分换成元）：和光一个聚合配成一对说
|
||||||
|
("sqlite", "SELECT SUM(amount) FROM orders", "SELECT SUM(amount) / 100 FROM orders",
|
||||||
|
["原流程算的是 SUM(amount)，盲算算的是 SUM(amount) / 100"]),
|
||||||
|
("sqlite", "SELECT SUM(amount) - SUM(fee) FROM orders", "SELECT SUM(amount), SUM(fee) FROM orders",
|
||||||
|
["原流程多算了 SUM(amount) - SUM(fee)，盲算没有"]),
|
||||||
|
], ids=["difference", "ratio", "window-partition", "scaled", "one-sided"])
|
||||||
|
def test_how_an_output_column_is_computed_is_compared(dialect, a, b, expected):
|
||||||
|
"""最外层算出来的列按算法比：两边用到的聚合一模一样（SUM(paid.total)、SUM(refunded.total)），只是谁减谁、谁除谁
|
||||||
|
换了，聚合那一面看不出来。"""
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b", [
|
||||||
|
("sqlite", PAID_REFUNDED + f"SELECT SUM(p.total) - SUM(r.total) {PR_JOIN}",
|
||||||
|
PAID_REFUNDED + "SELECT SUM(p.total) - SUM(r.total) FROM refunded r JOIN paid p ON p.user_id = r.user_id"),
|
||||||
|
("sqlite", "SELECT region, SUM(amount) * 1.0 / COUNT(*) AS avg_amt FROM orders GROUP BY region",
|
||||||
|
"SELECT o.region, 1.0 * SUM(o.amount) / COUNT(*) FROM orders o GROUP BY 1"),
|
||||||
|
("postgres", "SELECT order_id, RANK() OVER (PARTITION BY user_id ORDER BY amount DESC) AS k FROM orders",
|
||||||
|
"SELECT o.order_id, RANK() OVER (PARTITION BY o.user_id ORDER BY o.amount DESC) FROM orders o"),
|
||||||
|
# 最外层取整到分、空值记 0：还是那个合计
|
||||||
|
("sqlite", "SELECT SUM(amount) FROM orders", "SELECT ROUND(SUM(amount), 2) FROM orders"),
|
||||||
|
("mysql", "SELECT SUM(amount) FROM orders", "SELECT IFNULL(SUM(amount), 0) FROM orders"),
|
||||||
|
], ids=["swapped-join", "commutative", "window", "round", "default"])
|
||||||
|
def test_the_same_output_computation_written_differently_is_no_difference(dialect, a, b):
|
||||||
|
"""别名、加法乘法的先后、最外层取整、空值记 0 这类写法不同，算的是同一个数。"""
|
||||||
|
assert texts(a, b, dialect) == []
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b,expected", [
|
||||||
|
# 往里挪一层：net 的 v 是谁减谁
|
||||||
|
("sqlite",
|
||||||
|
PAID_REFUNDED + ", net AS (SELECT p.user_id, p.total - r.total AS v FROM paid p JOIN refunded r "
|
||||||
|
"ON r.user_id = p.user_id) SELECT SUM(v) FROM net",
|
||||||
|
PAID_REFUNDED + ", net AS (SELECT p.user_id, r.total - p.total AS v FROM paid p JOIN refunded r "
|
||||||
|
"ON r.user_id = p.user_id) SELECT SUM(v) FROM net",
|
||||||
|
["原流程的 net.v 是 paid.total - refunded.total，盲算是 refunded.total - paid.total"]),
|
||||||
|
("postgres",
|
||||||
|
PAID_REFUNDED + ", net AS (SELECT p.user_id, p.total AS v FROM paid p JOIN refunded r ON r.user_id = p.user_id) "
|
||||||
|
"SELECT COUNT(*) FROM net WHERE v > 100",
|
||||||
|
PAID_REFUNDED + ", net AS (SELECT p.user_id, r.total AS v FROM paid p JOIN refunded r ON r.user_id = p.user_id) "
|
||||||
|
"SELECT COUNT(*) FROM net WHERE v > 100",
|
||||||
|
["原流程的 net.v 是 paid.total，盲算是 refunded.total"]),
|
||||||
|
("mysql",
|
||||||
|
"SELECT SUM(x.total) FROM (SELECT p.user_id, p.total " + PR_SUBQUERIES + ") x",
|
||||||
|
"SELECT SUM(x.total) FROM (SELECT p.user_id, r.total " + PR_SUBQUERIES + ") x",
|
||||||
|
["原流程的 x.total 是 p.total，盲算是 r.total"]),
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, amount * 0.9 AS net FROM orders) SELECT SUM(net) FROM t",
|
||||||
|
"WITH t AS (SELECT user_id, amount * 0.8 AS net FROM orders) SELECT SUM(net) FROM t",
|
||||||
|
["原流程的 t.net 是 amount * 0.9，盲算是 amount * 0.8"]),
|
||||||
|
("mysql",
|
||||||
|
"WITH t AS (SELECT user_id, CASE WHEN status = 'paid' THEN amount ELSE 0 END AS paid_amt FROM orders) "
|
||||||
|
"SELECT SUM(paid_amt) FROM t",
|
||||||
|
"WITH t AS (SELECT user_id, CASE WHEN status IN ('paid', 'shipped') THEN amount ELSE 0 END AS paid_amt "
|
||||||
|
"FROM orders) SELECT SUM(paid_amt) FROM t",
|
||||||
|
["原流程的 t.paid_amt 是 CASE WHEN status = 'paid' THEN amount ELSE 0 END，"
|
||||||
|
"盲算是 CASE WHEN status IN ('paid', 'shipped') THEN amount ELSE 0 END"]),
|
||||||
|
("postgres", "SELECT COUNT(*) FROM (SELECT user_id, amount - fee AS net FROM orders) a WHERE a.net > 100",
|
||||||
|
"SELECT COUNT(*) FROM (SELECT user_id, amount + fee AS net FROM orders) b WHERE b.net > 100",
|
||||||
|
["原流程的 a.net 是 amount - fee，盲算的 b.net 是 amount + fee"]),
|
||||||
|
# 按支付月份还是下单月份分组
|
||||||
|
("mysql", "WITH t AS (SELECT DATE_FORMAT(paid_at, '%Y-%m') AS m, amount FROM orders) "
|
||||||
|
"SELECT m, SUM(amount) FROM t GROUP BY m",
|
||||||
|
"WITH t AS (SELECT DATE_FORMAT(created_at, '%Y-%m') AS m, amount FROM orders) "
|
||||||
|
"SELECT m, SUM(amount) FROM t GROUP BY m",
|
||||||
|
["原流程的 t.m 是 DATE_FORMAT(paid_at, '%Y-%m')，盲算是 DATE_FORMAT(created_at, '%Y-%m')"]),
|
||||||
|
("sqlite", "WITH t AS (SELECT strftime('%Y-%m', paid_at) AS m, amount FROM orders) "
|
||||||
|
"SELECT m, SUM(amount) FROM t GROUP BY m",
|
||||||
|
"WITH t AS (SELECT strftime('%Y-%m', created_at) AS m, amount FROM orders) "
|
||||||
|
"SELECT m, SUM(amount) FROM t GROUP BY m",
|
||||||
|
["原流程的 t.m 是 STRFTIME('%Y-%m', paid_at)，盲算是 STRFTIME('%Y-%m', created_at)"]),
|
||||||
|
# 窗口：每人第一笔还是最后一笔；有并列时 RANK 和 ROW_NUMBER 取到的行数不同
|
||||||
|
("postgres",
|
||||||
|
"SELECT SUM(amount) FROM (SELECT user_id, amount, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at) "
|
||||||
|
"AS rn FROM orders) t WHERE rn = 1",
|
||||||
|
"SELECT SUM(amount) FROM (SELECT user_id, amount, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at DESC) "
|
||||||
|
"AS rn FROM orders) t WHERE rn = 1",
|
||||||
|
["原流程的 t.rn 是 ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at)，"
|
||||||
|
"盲算是 ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at DESC)"]),
|
||||||
|
("mysql", "WITH r AS (SELECT user_id, amount, RANK() OVER (ORDER BY amount DESC) AS k FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM r WHERE k <= 10",
|
||||||
|
"WITH r AS (SELECT user_id, amount, ROW_NUMBER() OVER (ORDER BY amount DESC) AS k FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM r WHERE k <= 10",
|
||||||
|
["原流程的 r.k 是 RANK() OVER (ORDER BY amount DESC)，盲算是 ROW_NUMBER() OVER (ORDER BY amount DESC)"]),
|
||||||
|
# 外面直接 SELECT *：每一列都用到了
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, amount * 0.9 AS net FROM orders) SELECT * FROM t",
|
||||||
|
"WITH t AS (SELECT user_id, amount * 0.8 AS net FROM orders) SELECT * FROM t",
|
||||||
|
["原流程的 t.net 是 amount * 0.9，盲算是 amount * 0.8"]),
|
||||||
|
# 递归 CTE 自己写了列名：按列名找各支同一位置的定义
|
||||||
|
("postgres", "WITH RECURSIVE t(n) AS (SELECT 1 UNION ALL SELECT n + 1 FROM t WHERE n < 10) SELECT SUM(n) FROM t",
|
||||||
|
"WITH RECURSIVE t(n) AS (SELECT 1 UNION ALL SELECT n + 2 FROM t WHERE n < 10) SELECT SUM(n) FROM t",
|
||||||
|
["原流程的 t.n 是 1 | n + 1，盲算是 1 | n + 2"]),
|
||||||
|
], ids=["nested-difference", "nested-column", "nested-subquery", "scale", "case", "renamed-subquery", "month",
|
||||||
|
"month-strftime", "window-order", "window-function", "select-star", "recursive"])
|
||||||
|
def test_how_a_derived_column_is_computed_is_reported(dialect, a, b, expected):
|
||||||
|
"""外面写的都是 t.net、net.v、x.total，两边一样；它在 CTE、子查询里怎么算的不一样，要报出来，而且只报这一处：
|
||||||
|
外面按它写的条件、聚合两边写法一样，不跟着报。"""
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b", [
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, amount * 0.9 AS net FROM orders) SELECT SUM(net) FROM t",
|
||||||
|
"WITH x AS (SELECT user_id, 0.9 * amount AS net FROM orders) SELECT SUM(x.net) FROM x"),
|
||||||
|
# 外面没用到的列算法不同，不影响结果
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, amount, paid_at AS ts FROM orders) SELECT SUM(amount) FROM t",
|
||||||
|
"WITH t AS (SELECT user_id, amount, created_at AS ts FROM orders) SELECT SUM(amount) FROM t"),
|
||||||
|
# 取日期的两种写法
|
||||||
|
("postgres", "WITH t AS (SELECT DATE(paid_at) AS d, amount FROM orders) SELECT d, SUM(amount) FROM t GROUP BY d",
|
||||||
|
"WITH t AS (SELECT CAST(paid_at AS DATE) AS d, amount FROM orders) SELECT d, SUM(amount) FROM t GROUP BY d"),
|
||||||
|
("mysql",
|
||||||
|
"SELECT SUM(x.total) FROM (SELECT p.user_id, p.total " + PR_SUBQUERIES + ") x",
|
||||||
|
"SELECT SUM(y.total) FROM (SELECT a.user_id, a.total FROM (SELECT user_id, SUM(amount) AS total FROM refunds "
|
||||||
|
"GROUP BY user_id) b JOIN (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) a "
|
||||||
|
"ON a.user_id = b.user_id) y"),
|
||||||
|
], ids=["commutative", "unused-column", "date-cast", "nested-renamed"])
|
||||||
|
def test_the_same_derived_column_written_differently_is_no_difference(dialect, a, b):
|
||||||
|
"""派生来源里一列的写法不同、算的一样（0.9 * amount 和 amount * 0.9、DATE() 和 CAST(… AS DATE)），或者外面根本
|
||||||
|
没用到那一列：不报。"""
|
||||||
|
assert texts(a, b, dialect) == []
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b,expected", [
|
||||||
|
# 换了名字、算法一样：配成同一列
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t WHERE total > 100",
|
||||||
|
"WITH x AS (SELECT user_id, SUM(amount) AS sum_amt FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM x WHERE sum_amt > 100", []),
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, SUM(amount) AS total, COUNT(*) AS n FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t WHERE total > 100 AND n > 1",
|
||||||
|
"WITH t AS (SELECT user_id, SUM(amount) AS s, COUNT(*) AS c FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t WHERE s > 100 AND c > 1", []),
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, amount, DATE(paid_at) AS d FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'",
|
||||||
|
"WITH t AS (SELECT user_id, amount, DATE(paid_at) AS day FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE day >= '2026-09-01'", []),
|
||||||
|
# 换了名字又改了算法：两边各剩一列，配成同一列，外面的条件不报成两边各有一个
|
||||||
|
("sqlite", "WITH t AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t WHERE total > 100",
|
||||||
|
"WITH x AS (SELECT user_id, SUM(amount - discount) AS net_total FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM x WHERE net_total > 100",
|
||||||
|
["原流程对 amount 求和，盲算对 amount - discount 求和",
|
||||||
|
"原流程的 t.total 是 SUM(amount)，盲算的 x.net_total 是 SUM(amount - discount)"]),
|
||||||
|
], ids=["renamed", "two-renamed", "renamed-date", "renamed-and-changed"])
|
||||||
|
def test_derived_columns_renamed_on_one_side_are_paired(dialect, a, b, expected):
|
||||||
|
"""CTE、子查询选出来的列外面按名字引用，盲算常给它另起名字（total、sum_amt）：按定义配成同一列，外面按它写的条件
|
||||||
|
不算两边各有一个。"""
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
# diff_facets：时间列写在 CTE、子查询里
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect", DIALECTS)
|
||||||
|
def test_a_time_column_computed_in_a_cte_is_seen_through(dialect):
|
||||||
|
"""按支付时间还是下单时间筛，日期表达式写在 CTE 里也认得出：时间那一面透过 d = DATE(paid_at) 认出是 paid_at，带上
|
||||||
|
表名、列名给调用方对照数据目录；d 的定义不同不再另报一遍。"""
|
||||||
|
a = ("WITH t AS (SELECT user_id, amount, DATE(paid_at) AS d FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'")
|
||||||
|
b = a.replace("DATE(paid_at)", "DATE(created_at)")
|
||||||
|
[diff] = diff_facets(facets(a, dialect), facets(b, dialect), labels=LABELS,
|
||||||
|
names={"orders.paid_at": "支付时间", "created_at": "下单时间"})
|
||||||
|
assert (diff.aspect, diff.text) == ("time", "原流程按支付时间 paid_at 筛时间，盲算按下单时间 created_at")
|
||||||
|
assert diff.detail == {"a": [{"table": "orders", "column": "paid_at"}],
|
||||||
|
"b": [{"table": "orders", "column": "created_at"}]}
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b,expected", [
|
||||||
|
("postgres", "SELECT SUM(x.amount) FROM (SELECT amount, CAST(paid_at AS DATE) AS day FROM orders) x "
|
||||||
|
"WHERE x.day BETWEEN '2026-09-01' AND '2026-09-30'",
|
||||||
|
"SELECT SUM(y.amount) FROM (SELECT amount, CAST(created_at AS DATE) AS day FROM orders) y "
|
||||||
|
"WHERE y.day BETWEEN '2026-09-01' AND '2026-09-30'",
|
||||||
|
["原流程按 paid_at 筛时间，盲算按 created_at"]),
|
||||||
|
# 两个 CTE 同名、定义对调：原流程筛付款时间，盲算筛退款时间
|
||||||
|
("mysql", "WITH a AS (SELECT user_id, MAX(paid_at) AS ts FROM orders GROUP BY user_id), "
|
||||||
|
"b AS (SELECT user_id, MAX(refunded_at) AS ts FROM refunds GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM a JOIN b ON b.user_id = a.user_id WHERE a.ts >= '2026-09-01'",
|
||||||
|
"WITH a AS (SELECT user_id, MAX(refunded_at) AS ts FROM refunds GROUP BY user_id), "
|
||||||
|
"b AS (SELECT user_id, MAX(paid_at) AS ts FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM a JOIN b ON b.user_id = a.user_id WHERE a.ts >= '2026-09-01'",
|
||||||
|
["原流程按 paid_at 筛时间，盲算按 refunded_at"]),
|
||||||
|
# 带时区偏移的不是同一天：时间列一样，d 的定义不同照报
|
||||||
|
("sqlite", "WITH t AS (SELECT amount, DATE(paid_at) AS d FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'",
|
||||||
|
"WITH t AS (SELECT amount, DATE(paid_at, '+8 hours') AS d FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'",
|
||||||
|
["原流程的 t.d 是 DATE(paid_at)，盲算是 DATE(paid_at, '+8 hours')"]),
|
||||||
|
("postgres", "WITH t AS (SELECT DATE(paid_at) AS d, amount FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'",
|
||||||
|
"WITH t AS (SELECT CAST(paid_at AS DATE) AS d, amount FROM orders) "
|
||||||
|
"SELECT SUM(amount) FROM t WHERE d >= '2026-09-01'", []),
|
||||||
|
], ids=["cast-between", "same-name-swapped", "time-zone", "date-or-cast"])
|
||||||
|
def test_time_columns_inside_derived_tables(dialect, a, b, expected):
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
# diff_facets：两边写法一样、意思不同时说清楚
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
|
||||||
|
|
||||||
|
def test_same_named_derived_tables_with_swapped_definitions_are_told_apart():
|
||||||
|
"""两边的 CTE 同名、定义对调了（盲算的 t1 是原流程的 t2）：按结构配对之后 t1.total 两边不是同一列，可只写
|
||||||
|
「t1.total > 100」两边一模一样，自相矛盾。带上各自的来历：原流程筛付款合计，盲算筛退款合计。"""
|
||||||
|
a = ("WITH t1 AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id), "
|
||||||
|
"t2 AS (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t1 JOIN t2 ON t2.user_id = t1.user_id WHERE t1.total > 100")
|
||||||
|
b = ("WITH t1 AS (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id), "
|
||||||
|
"t2 AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM t1 JOIN t2 ON t2.user_id = t1.user_id WHERE t1.total > 100")
|
||||||
|
assert texts(a, b, "postgres", labels=LABELS) == [
|
||||||
|
"原流程按 t1.total > 100（t1.total 即 SUM(orders.amount)）筛选，"
|
||||||
|
"盲算按 t1.total > 100（t1.total 即 SUM(refunds.amount)）筛选"]
|
||||||
|
a_sum = ("WITH a AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id), "
|
||||||
|
"b AS (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id) "
|
||||||
|
"SELECT SUM(a.total) FROM a JOIN b ON b.user_id = a.user_id")
|
||||||
|
b_sum = ("WITH a AS (SELECT user_id, SUM(amount) AS total FROM refunds GROUP BY user_id), "
|
||||||
|
"b AS (SELECT user_id, SUM(amount) AS total FROM orders GROUP BY user_id) "
|
||||||
|
"SELECT SUM(a.total) FROM a JOIN b ON b.user_id = a.user_id")
|
||||||
|
assert texts(a_sum, b_sum, labels=LABELS) == [
|
||||||
|
"原流程对 a.total（即 SUM(orders.amount)）求和，盲算对 a.total（即 SUM(refunds.amount)）求和"]
|
||||||
|
|
||||||
|
|
||||||
|
def test_same_named_derived_tables_that_differ_only_in_conditions_say_which_is_which():
|
||||||
|
"""两个 CTE 读同一张表、各列定义一样，只差里面的条件；盲算把两个名字的定义对调了、还都多了 is_deleted = 0。按条件
|
||||||
|
的重合配（不按名字：盲算的 paid 筛的是 status = 'open'），外面的「> 100」筛的就是未付合计；来历也一样（都是
|
||||||
|
SUM(orders.amount)），说清两边同名的各对应哪一个。"""
|
||||||
|
a = ("WITH paid AS (SELECT user_id, SUM(amount) AS total FROM orders WHERE status = 'paid' GROUP BY user_id), "
|
||||||
|
"open AS (SELECT user_id, SUM(amount) AS total FROM orders WHERE status = 'open' GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM paid p JOIN open o ON o.user_id = p.user_id WHERE p.total > 100")
|
||||||
|
b = ("WITH paid AS (SELECT user_id, SUM(amount) AS total FROM orders WHERE status = 'open' AND is_deleted = 0 "
|
||||||
|
"GROUP BY user_id), "
|
||||||
|
"open AS (SELECT user_id, SUM(amount) AS total FROM orders WHERE status = 'paid' AND is_deleted = 0 "
|
||||||
|
"GROUP BY user_id) "
|
||||||
|
"SELECT COUNT(*) FROM paid p JOIN open o ON o.user_id = p.user_id WHERE p.total > 100")
|
||||||
|
assert texts(a, b, labels=LABELS) == [
|
||||||
|
"原流程按 paid.total > 100 筛选，"
|
||||||
|
"盲算按 paid.total > 100（盲算的 paid 对应原流程的 open、盲算的 open 对应原流程的 paid）筛选",
|
||||||
|
"盲算只保留 is_deleted = 0 的行，原流程没有这个条件"]
|
||||||
|
|
||||||
|
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
# diff_facets：同一个条件筛的先后
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b,expected", [
|
||||||
|
# 「第一单就是已付的人数」对「有已付订单的人数」
|
||||||
|
("postgres",
|
||||||
|
"SELECT COUNT(*) FROM (SELECT user_id, status, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at) AS rn "
|
||||||
|
"FROM orders) t WHERE t.rn = 1 AND t.status = 'paid'",
|
||||||
|
"SELECT COUNT(*) FROM (SELECT user_id, status, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at) AS rn "
|
||||||
|
"FROM orders WHERE status = 'paid') t WHERE t.rn = 1",
|
||||||
|
["原流程先算 t 的窗口函数、再按 status = 'paid' 筛选，盲算先按 status = 'paid' 筛选、再算 t 的窗口函数"]),
|
||||||
|
("sqlite", "WITH t AS (SELECT * FROM orders ORDER BY amount DESC LIMIT 10) SELECT SUM(amount) FROM t "
|
||||||
|
"WHERE status = 'paid'",
|
||||||
|
"WITH t AS (SELECT * FROM orders WHERE status = 'paid' ORDER BY amount DESC LIMIT 10) SELECT SUM(amount) FROM t",
|
||||||
|
["原流程先取 t 的前 10 行、再按 status = 'paid' 筛选，盲算先按 status = 'paid' 筛选、再取 t 的前 10 行"]),
|
||||||
|
("sqlite", "WITH t AS (SELECT * FROM orders ORDER BY amount DESC LIMIT 10) SELECT SUM(amount) FROM t "
|
||||||
|
"WHERE paid_at >= '2026-09-01'",
|
||||||
|
"WITH t AS (SELECT * FROM orders WHERE paid_at >= '2026-09-01' ORDER BY amount DESC LIMIT 10) "
|
||||||
|
"SELECT SUM(amount) FROM t",
|
||||||
|
["原流程先取 t 的前 10 行、再按 paid_at >= '2026-09-01' 筛选，"
|
||||||
|
"盲算先按 paid_at >= '2026-09-01' 筛选、再取 t 的前 10 行"]),
|
||||||
|
], ids=["window", "limit", "limit-time"])
|
||||||
|
def test_filtering_before_or_after_a_window_or_limit_is_reported(dialect, a, b, expected):
|
||||||
|
"""先取前 10 行再筛已付，和先筛已付再取前 10 行，取到的不是同一批：派生来源取前几行、算窗口函数时，外面的条件
|
||||||
|
不再透成里面那张表的条件，两边只差在筛的先后，照报。"""
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b", [
|
||||||
|
# 按窗口的分区键筛，窗口里外都一样
|
||||||
|
("postgres",
|
||||||
|
"SELECT COUNT(*) FROM (SELECT user_id, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at) AS rn "
|
||||||
|
"FROM orders) t WHERE t.rn = 1 AND t.user_id > 100",
|
||||||
|
"SELECT COUNT(*) FROM (SELECT user_id, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY paid_at) AS rn "
|
||||||
|
"FROM orders WHERE user_id > 100) t WHERE t.rn = 1"),
|
||||||
|
# 两边都是先取前 10 行再筛，只是起的名字不同
|
||||||
|
("sqlite", "WITH t AS (SELECT * FROM orders ORDER BY amount DESC LIMIT 10) SELECT SUM(amount) FROM t "
|
||||||
|
"WHERE status = 'paid'",
|
||||||
|
"WITH top10 AS (SELECT * FROM orders ORDER BY amount DESC LIMIT 10) SELECT SUM(t.amount) FROM top10 t "
|
||||||
|
"WHERE t.status = 'paid'"),
|
||||||
|
], ids=["partition-key", "same-order"])
|
||||||
|
def test_filtering_at_an_equivalent_stage_is_no_difference(dialect, a, b):
|
||||||
|
assert texts(a, b, dialect) == []
|
||||||
|
|
||||||
|
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
# diff_facets：没用到的 CTE
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
|
||||||
|
|
||||||
|
def test_a_cte_defined_but_not_used_does_not_count():
|
||||||
|
"""盲算定义了只筛已付的 CTE，却直接查 orders：少的那个条件要报出来；定义了没用的 CTE 里多一个条件、多一张表，
|
||||||
|
不影响结果，不报。"""
|
||||||
|
paid = "WITH paid AS (SELECT * FROM orders WHERE status = 'paid') "
|
||||||
|
assert texts(paid + "SELECT SUM(amount) FROM paid", paid + "SELECT SUM(amount) FROM orders", labels=LABELS) == \
|
||||||
|
["原流程只保留 status = 'paid' 的行，盲算没有这个条件"]
|
||||||
|
assert texts("SELECT SUM(amount) FROM orders WHERE status = 'paid'",
|
||||||
|
"WITH junk AS (SELECT * FROM users WHERE vip = 1) "
|
||||||
|
"SELECT SUM(amount) FROM orders WHERE status = 'paid'") == []
|
||||||
|
assert texts(paid + "SELECT SUM(amount) FROM paid",
|
||||||
|
paid + "SELECT SUM(amount) FROM orders WHERE status = 'paid'") == []
|
||||||
|
|
||||||
|
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
# diff_facets：自己关联自己
|
||||||
|
# --------------------------------------------------------------------------
|
||||||
|
|
||||||
|
SELF_BASE = ("WITH base AS (SELECT user_id, referrer_id, SUM(amount) AS total FROM orders "
|
||||||
|
"GROUP BY user_id, referrer_id) ")
|
||||||
|
|
||||||
|
|
||||||
|
@pytest.mark.parametrize("dialect,a,b,expected", [
|
||||||
|
("postgres", SELF_BASE + "SELECT COUNT(*) FROM base b1 JOIN base b2 ON b2.user_id = b1.referrer_id "
|
||||||
|
"WHERE b1.total > b2.total",
|
||||||
|
SELF_BASE + "SELECT COUNT(*) FROM base b1 JOIN base b2 ON b2.user_id = b1.referrer_id WHERE b2.total > b1.total",
|
||||||
|
["原流程只保留 b1.total > b2.total 的行，盲算没有这个条件", "盲算只保留 b1.total < b2.total 的行，原流程没有这个条件"]),
|
||||||
|
("sqlite", "SELECT COUNT(*) FROM orders o1 JOIN orders o2 ON o2.id = o1.parent_id WHERE o1.amount > o2.amount",
|
||||||
|
"SELECT COUNT(*) FROM orders o1 JOIN orders o2 ON o2.id = o1.parent_id WHERE o2.amount > o1.amount",
|
||||||
|
["原流程只保留 o2.amount < o1.amount 的行，盲算没有这个条件", "盲算只保留 o2.amount > o1.amount 的行，原流程没有这个条件"]),
|
||||||
|
# 换了别名、换了关联先后：b1 还是出 referrer_id 的那一个
|
||||||
|
("postgres", SELF_BASE + "SELECT COUNT(*) FROM base b1 JOIN base b2 ON b2.user_id = b1.referrer_id "
|
||||||
|
"WHERE b1.total > b2.total",
|
||||||
|
SELF_BASE + "SELECT COUNT(*) FROM base y JOIN base x ON x.referrer_id = y.user_id WHERE y.total < x.total", []),
|
||||||
|
("sqlite", "SELECT COUNT(*) FROM orders o1 JOIN orders o2 ON o2.id = o1.parent_id WHERE o1.amount > o2.amount",
|
||||||
|
"SELECT COUNT(*) FROM orders c JOIN orders p ON c.parent_id = p.id WHERE c.amount > p.amount", []),
|
||||||
|
], ids=["cte-reversed", "table-reversed", "cte-renamed", "table-renamed"])
|
||||||
|
def test_a_self_join_tells_its_two_aliases_apart(dialect, a, b, expected):
|
||||||
|
"""同一个 CTE、同一张表自己关联自己：两个别名按各自在关联条件里出的列区分（出 referrer_id 的、出 user_id 的），
|
||||||
|
比较写反了要报；只是换了别名、换了关联先后不报。"""
|
||||||
|
assert texts(a, b, dialect, labels=LABELS) == expected
|
||||||
|
|
||||||
|
|
||||||
@pytest.mark.parametrize("dialect", DIALECTS)
|
@pytest.mark.parametrize("dialect", DIALECTS)
|
||||||
def test_each_dialect_gives_the_same_three_findings(dialect):
|
def test_each_dialect_gives_the_same_three_findings(dialect):
|
||||||
q = {"sqlite": '"', "postgres": '"', "mysql": "`"}[dialect]
|
q = {"sqlite": '"', "postgres": '"', "mysql": "`"}[dialect]
|
||||||
|
|||||||
Reference in New Issue
Block a user