我有一个复杂的 SQL 查询,可以成功提取组件产品记录的成本并计算父/捆绑产品的总体成本。当每个组件都有供应商成本并且本身不是父/捆绑产品时,这种方法就有效。
查询#1
SET @parentid = 36; SELECT sub.product_sku AS product_sku, sub.product_label AS product_label, c2p.bundle_parentid AS bundle_parentid, c2p.componentid AS comp_product_id, sub.qty AS qty, sub.preferred AS preferred, sub.supply_qty AS supply_qty, sub.cost AS cost, ROUND((SELECT cost)/(SELECT supply_qty),2) AS adjusted_cost FROM products AS p JOIN component2bundle AS c2p ON c2p.componentid = p.product_id JOIN ( /* Get the preferred/cheapest supplier date for this component */ SELECT p2.product_sku AS product_sku, p2.product_label AS product_label, IFNULL(s2p2.cost, NULL) AS cost, s2p2.productid AS product_id, s2p2.supplier_preferred AS preferred, s2p2.product_quantity AS supply_qty, c2p2.componentid AS comp_product_id, c2p2.component_quantity AS qty, c2p2.bundle_parentid AS bundle_parentid FROM products AS p2 INNER JOIN supplier2product AS s2p2 ON s2p2.productid = p2.product_id INNER JOIN component2bundle AS c2p2 ON s2p2.productid = c2p2.componentid WHERE c2p2.bundle_parentid = @parentid AND c2p2.c2p_archive = 0 AND COALESCE(s2p2.s2p_archive,0) = 0 ORDER BY c2p2.componentid ASC, s2p2.supplier_preferred DESC, s2p2.cost ASC ) AS sub ON (sub.product_id = c2p.componentid) WHERE c2p.bundle_parentid = @parentid;
我的目标是调整或重写查询,以便它可以消除任何捆绑组件的成本,因此递归 CTE 查询似乎是继续的方法。
我已经成功编写了一个 CTE 查询,该查询可以从显示父 -> 子关系的表中提取每个组件产品 ID,并为每个组件分配层次结构中的级别。我正在努力解决的是如何将两者整合起来。
CTE查询
WITH RECURSIVE components AS ( SELECT componentid, 1 AS level FROM component2bundle WHERE bundle_parentid = 'target_productID' UNION ALL SELECT c2b.componentid, c.level+1 FROM components c, component2bundle c2b WHERE c2b.bundle_parentid = c.componentid ) SELECT * FROM components ORDER BY level DESC;
我在这里创建了一个 MySQL 8.0 fiddle 以帮助提供更好的上下文:
https://dbfiddle.uk/M6HT_R13
注意:我已经削减了查询#1,使其更容易处理,因此可以忽略小提琴中的某些字段。
*编辑:在fiddle中设置parentid变量以查看当前查询如何拉取:
一些附加说明。
查询#1中的子查询旨在从supplier2cost表中提取首选或(如果未设置)最低的供应商成本,而我不确定如何在其中实现这个子查询CTE 上下文(如果有的话)。
如果其他上下文有帮助,请询问,我将编辑查询以提供该信息。
预期/预期输出
ProductSKU | 产品标签 | BundleParentID | Component_ID | 级别 | 数量 | 首选 | 供应数量 | 成本 | 调整后的成本 |
---|---|---|---|---|---|---|---|---|---|
子组件#1 | CMP#2 | 36 | 35 | 2 | 1 | 1 | 1 | 费用 | 每单位成本 |
子组件#2 | CMP#3 | 36 | 37 | 2 | 1 | 1 | 1 | 费用 | 每单位成本 |
子组件#3 | CMP#4 | 36 | 38 | 2 | 1 | 1 | 1 | 费用 | 每单位成本 |
组件#1 | CMP#1 | 34 | 33 | 1 | 1 | 1 | 1 | 费用 | 每单位成本 |
子包 | 外滩#1 | 36 | 33 | 1 | 1 | 1 | 1 | 费用 | 每单位成本 |
数据最终将用于提供如下组件成本表:
P粉6752585982024-01-17 10:16:36
您可能想要这样的东西:
在第二个 cte
上,添加了一个条件,将您的主体查询与递归查询连接起来,以仅提取选定的查询
SET @parentid = 34; WITH RECURSIVE components AS ( SELECT componentid, p.product_sku, 1 AS level FROM component2bundle JOIN products p ON componentid = p.product_id WHERE bundle_parentid = @parentid UNION ALL SELECT c2b.componentid, product_sku, c.level+1 FROM components c, component2bundle c2b WHERE c2b.bundle_parentid = c.componentid ), CTE AS ( SELECT sub.product_sku AS product_sku, sub.product_label AS product_label, c2p.bundle_parentid AS bundle_parentid, c2p.componentid AS comp_product_id, sub.qty AS qty, sub.preferred AS preferred, sub.supply_qty AS supply_qty, sub.cost AS cost, ROUND((SELECT cost)/(SELECT supply_qty),2) AS adjusted_cost, c.level FROM products AS p JOIN component2bundle AS c2p ON c2p.componentid = p.product_id JOIN components c on c.componentid = c2p.componentid JOIN ( /* Get the preferred/cheapest supplier date for this component */ SELECT p2.product_sku AS product_sku, p2.product_label AS product_label, IFNULL(s2p2.cost, NULL) AS cost, s2p2.productid AS product_id, s2p2.supplier_preferred AS preferred, s2p2.product_quantity AS supply_qty, c2p2.componentid AS comp_product_id, c2p2.component_quantity AS qty, c2p2.bundle_parentid AS bundle_parentid FROM products AS p2 INNER JOIN supplier2product AS s2p2 ON s2p2.productid = p2.product_id INNER JOIN component2bundle AS c2p2 ON s2p2.productid = c2p2.componentid WHERE c2p2.c2p_archive = 0 AND COALESCE(s2p2.s2p_archive,0) = 0 ORDER BY c2p2.componentid ASC, s2p2.supplier_preferred DESC, s2p2.cost ASC ) AS sub ON (sub.product_id = c2p.componentid) ) SELECT * FROM CTE c WHERE preferred = 1