如何模拟依赖 CTE答案

【问题标题】：How to emulate a dependent CTE如何模拟依赖 CTE
【发布时间】：2022-06-12 02:36:46
【问题描述】：

从这个问题Possible to emulate a basic CTE in MySQL by in-lining it 跟进，是否可以在不使用 CTE 的情况下在 mysql5.7 中内联以下查询？

WITH tbl1 AS (
    SELECT 1 AS a
), tbl2 AS (
    SELECT 1 AS a UNION ALL SELECT 2
), tbl3 AS (
    SELECT * FROM tbl1 JOIN tbl2 USING (a)
) SELECT * FROM tbl3, tbl2

如果是这样，那怎么做？

【问题讨论】：

如果您的 CTE 中没有非确定性构造，或者如果此类 CTE 使用过一次，则只需将 CTE 名称嵌套地替换为其子查询。如果存在非确定性构造，并且必须多次使用 CTE，则为这些 CTE 临时创建（不是临时！）表是唯一的选择。
@Akina 你能否澄清你的意思：then temporarily created (not temporary!) tables for these CTEs is the only option。你将如何为WITH cte AS (SELECT RAND() AS a)... 创建一个临时创建的表？
作为单独的CREATE TABLE .. SELECT 查询。
@Akina 当然，但这与 tmp 表有什么不同？无论如何，它会在使用后被销毁，所以我认为这不是问题。
临时表不能在查询中多次使用。

标签： mysql common-table-expression

【解决方案1】：

这样做没有什么特别的问题，只是你必须重复 tbl2：

SELECT *
FROM (
    SELECT *
    FROM (
        SELECT 1 AS a
    ) tbl1
    JOIN (
        SELECT 1 AS a UNION ALL SELECT 2
    ) tbl2
    USING (a)
) tbl3
CROSS JOIN
(
    SELECT 1 AS a UNION ALL SELECT 2
) tbl2

fiddle

【讨论】：

【解决方案2】：

这类似于替换函数的主体。以类似的方式编写它（从概念上讲）：

let tbl1 =  (SELECT 1 AS a)                          AS tbl1
let tbl2 =  (SELECT 1 AS a UNION ALL SELECT 2)       AS tbl2
let tbl3 =  (SELECT * FROM tbl1 JOIN tbl2 USING (a)) AS tbl3
-- rewriting tbl1 for tbl1 and tbl2
let tbl3 =  (SELECT * FROM (SELECT 1 AS a) AS tbl1 
             JOIN ((SELECT 1 AS a UNION ALL SELECT 2)) AS tbl1 USING (a)) AS tbl3


SELECT * FROM tbl3, tbl2
-- substitutie for tbl3, tbl2
SELECT * FROM 
    (SELECT * FROM (SELECT 1 AS a) AS tbl1 
              JOIN ((SELECT 1 AS a UNION ALL SELECT 2)) AS tbl1 USING (a)) AS tbl3,
    (SELECT 1 AS a UNION ALL SELECT 2) AS tbl2

【讨论】：