Oracle CTE递归实现PCB行业的叠层关系

1、需求背景，出货报告要实现叠板假层的处理，需求如下

表ID,layer,MEDIUM数据如下
第一种情况，layer有K的

ID	layer	MEDIUM
1	L1-L2	30
2	L2-L3	40
3	L3-K1	20
4	K1-L4	10
5	L4-L5	20
6	L5-L6	30
7	L7-K2	10
8	K2-L8	11
9	L8-L9	10
10	L9-L10	30

实现layer有K1的，L3-L4,并合计 MEDIUM列等于30，layer有K2的，L7-L8,并合计 MEDIUM列等于21

layer	MEDIUM
L1-L2	30
L2-L3	40
L3-L4	30
L4-L5	20
L5-L6	30
L7-L8	21
L8-L9	10
L9-L10	30

第二种情况，layer有K的

ID	layer	MEDIUM
1	L1-L2	30
2	L2-L3	40
3	L3-K1	20
4	K1-K2	10
5	K2-L4	20
6	L4-L5	30
7	L5-L6	10

实现layer有K的，L3-L4,并合计 MEDIUM列等于50=(20+10+20)

layer	MEDIUM
L1-L2	30
L2-L3	40
L3-L4	50
L4-L5	30
L5-L6	10

处理思路

该方案通过递归CTE构建完整路径链，自动处理单K节点（如L3-K1-L4）和多K连续节点（如L3-K1-K2-L4）两种情况，输出合并后的层级路径及MEDIUM总和。

执行示例：

第一种情况输出：

L3-L4 | 30
L7-L8 | 21

第二种情况输出：

L3-L4 | 50 (20+10+20)

关键特性：

动态节点识别：自动检测所有K节点（K1/K2等）15
路径完整性检查：确保合并后的路径始终以非K节点结尾68
最短路径优先：当存在多条合并路径时选择最短链路

WITH 表达式

WITH split_data AS (SELECT ID,layer,MEDIUM,REGEXP_SUBSTR(layer, '[^-]+') AS start_node,REGEXP_SUBSTR(layer, '[^-]+$') AS end_nodeFROM layer_data
),
path_chains AS (-- 基础路径（起始节点非K的路径）SELECT ID, layer, MEDIUM,start_node AS chain_start,end_node AS chain_end,CAST(layer AS VARCHAR2(4000)) AS full_path,MEDIUM AS total_medium,1 AS path_levelFROM split_data WHERE start_node NOT LIKE 'K%'UNION ALL-- 递归连接后续路径SELECT d.ID, d.layer, d.MEDIUM,p.chain_start,d.end_node AS chain_end,p.full_path || '→' || d.layer,p.total_medium + d.MEDIUM,p.path_level + 1FROM path_chains pJOIN split_data d ON p.chain_end = d.start_nodeWHERE p.chain_end NOT LIKE 'L%'  -- 仅连接K节点或中间节点
),
k_merged AS (SELECT chain_start || '-' || chain_end AS combined_layer,total_medium,ROW_NUMBER() OVER (PARTITION BY chain_start, chain_end ORDER BY path_level) AS rnFROM path_chainsWHERE chain_end LIKE 'L%'  -- 最终结束节点需为非K节点AND full_path LIKE '%K%'   -- 必须包含K节点
)
SELECT combined_layer,total_medium
FROM k_merged
WHERE rn = 1  -- 取最短路径合并结果
ORDER BY combined_layer;