SQL樹形路徑結構長度

Question

我有以下的值：

Id | Sum | Tree_path_structure 
1 | 10 | /1/2 
2 | 30 | /1/3/4 
3 | 40 | /1/3/5 
4 | 50 | /1/6 

對於最長樹的路徑，在該示例中，在ID（2,3）的查詢應該返回小數目和輸出應該是：

Id | Sum | Tree_path_structure 
1 | 10 | /1/2 
2 | 30 | /1/3/4 
4 | 50 | /1/6 

我想基於樹路徑結構長度的SQL將返回最小的總和值。

Answer 1

甲骨文設置：

CREATE TABLE your_table (Id, SUm, Tree_path_structure) AS 
    SELECT 1, 10, '/1/2' FROM DUAL UNION ALL 
    SELECT 2, 30, '/1/3/4' FROM DUAL UNION ALL 
    SELECT 3, 40, '/1/3/5' FROM DUAL UNION ALL 
    SELECT 4, 50, '/1/6' FROM DUAL; 

查詢：

SELECT id, 
     sum, 
     tree_path_structure 
FROM (
    SELECT id, 
     sum, 
     tree_path_structure, 
     CASE 
      WHEN is_Max_Depth = 0 
      OR sum = MIN(sum) OVER (PARTITION BY is_Max_Depth) 
      THEN 0 
      ELSE 1 
     END AS is_Max_Depth_With_Higher_sum 
    FROM (
    SELECT id, 
      sum, 
      tree_path_structure, 
      CASE depth WHEN MAX(depth) OVER() THEN 1 ELSE 0 END AS is_Max_Depth 
    FROM (
     SELECT ID, 
      SUM, 
      Tree_Path_Structure, 
      LENGTH(Tree_Path_Structure) - LENGTH(REPLACE(Tree_Path_Structure, '/')) 
       AS depth 
     FROM your_table 
    ) t 
) 
) 
WHERE is_Max_Depth_With_Higher_sum = 0; 

輸出：

 ID  SUM TREE_P 
---------- ---------- ------ 
     1   10 /1/2 
     4   50 /1/6 
     2   30 /1/3/4 

更新：

SELECT id, 
     sum, 
     tree_path_structure 
FROM (
    SELECT id, 
     sum, 
     tree_path_structure, 
     CASE 
      WHEN is_Max_Depth = 0 
      OR sum = MIN(sum) OVER (PARTITION BY root, is_Max_Depth) 
      THEN 0 
      ELSE 1 
     END AS is_Max_Depth_With_Higher_sum 
    FROM (
    SELECT id, 
      sum, 
      tree_path_structure, 
      root, 
      CASE depth WHEN MAX(depth) OVER (PARTITION BY root) THEN 1 ELSE 0 END AS is_Max_Depth 
    FROM (
     SELECT ID, 
      SUM, 
      Tree_Path_Structure, 
      SUBSTR(Tree_Path_Structure, 1, INSTR(Tree_Path_Structure, '/', 1, 2)) 
       AS root, 
      LENGTH(Tree_Path_Structure) - LENGTH(REPLACE(Tree_Path_Structure, '/')) 
       AS depth 
     FROM your_table 
    ) t 
) 
) 
WHERE is_Max_Depth_With_Higher_sum = 0;