如何获得层表

Question

我一直在挣扎了一下关于如何处理这种情况的路径：

我有一个表结构如下：

Family_code | Parent_Family_Code | .... 
    1     2 
    2     4 
    3     6 
    4     3 
    ...................... 

当用户搜索特定家族码，我需要返回的整个路径（最多10个级别最大），所以例如用于family_code = 1，我需要：

Family_code | parent_1 | p_2 | p_3 | p_4 | p_5 | ..... 
     1   2  4  3  6  null null..... 

我知道我可以使用sys_connect_by_path()这将带给我预期的结果，但作为一个字符串，而不是单独的列，这是我宁愿避免的。

这也可以通过10个左连接来完成同一个表，或者使用LEAD()/LAG()函数，这些函数将包含大量的子查询，并且会造成凌乱和难以理解的查询，但是再一次，这将会更加沉重那么它应该是，我需要尽可能简化它。

我想出了使用substr()功能的解决方案（代码的长度将永远是VARCHAR2（3））：

SELECT s.family_code, 
s.parent_family_code_1, 
s.parent_family_code_2, 
CASE WHEN length(s.family_path) - (4 * 3 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 3 + 2), 3) ELSE NULL END as parent_family_code_3, 
CASE WHEN length(s.family_path) - (4 * 4 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 4 + 2), 3) ELSE NULL END as parent_family_code_4, 
CASE WHEN length(s.family_path) - (4 * 5 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 5 + 2), 3) ELSE NULL END as parent_family_code_5, 
CASE WHEN length(s.family_path) - (4 * 6 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 6 + 2), 3) ELSE NULL END as parent_family_code_6, 
CASE WHEN length(s.family_path) - (4 * 7 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 7 + 2), 3) ELSE NULL END as parent_family_code_7, 
CASE WHEN length(s.family_path) - (4 * 8 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 8 + 2), 3) ELSE NULL END as parent_family_code_8, 
CASE WHEN length(s.family_path) - (4 * 9 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 9 + 2), 3) ELSE NULL END as parent_family_code_9, 
CASE WHEN length(s.family_path) - (4 * 10 + 2) > 0 THEN substr(s.family_path, length(s.family_path) - (4 * 10 + 2), 3) ELSE NULL END as parent_family_code_10 
    FROM (SELECT t.family_code, 
       t.parent_family_code as parent_family_code_1, 
       prior t.parent_family_code as parent_family_code_2, 
       sys_connect_by_path(t.family_code, ',') as family_path 
      FROM table t 
     connect by prior t.family_code = t.parent_family_code) s 

但我想没有，因为它使用的子串的解决方案当其他开发人员碰到它时，将很难做任何修改。 。 所以基本上我的问题是 - 如何在不使用子串的情况下选择整个路径作为不同的列？

Answer 1

稍微修改的查询从@MT0回答，使用PIVOT条款。

SELECT * 
FROM (
    select connect_by_root(family_code) as Family_code, 
      'P_' || level lev_el, 
      parent_family_code 
    from table_name t 
    start with not exists(
     select 1 from table_name t1 
     where t.family_code = t1.parent_family_code) 
    connect by prior parent_family_code = family_code 
) 
PIVOT (
    max(parent_family_code) 
    FOR (lev_el) IN ( 
     'P_1', 'P_2', 'P_3', 'P_4', 'P_5', 'P_6','P_7', 'P_8','P_9','P_10' , 
     'P_11', 'P_12', 'P_13', 'P_14', 'P_15', 'P_16','P_17', 'P_18','P_19','P_20', 
     'P_21', 'P_22', 'P_23', 'P_24', 'P_25', 'P_26','P_27', 'P_28','P_29','P_30' 
     /* add more "levels" here if required */ 
) 
); 

查询的样本数据来自@ MT0回答结果（@ MT0，感谢您提供的样本数据）：

FAMILY_CODE  'P_1'  'P_2'  'P_3'  'P_4'  'P_5'  'P_6'  'P_7'  'P_8'  'P_9'  'P_10'  'P_11'  'P_12'  'P_13'  'P_14'  'P_15'  'P_16'  'P_17'  'P_18'  'P_19'  'P_20'  'P_21'  'P_22'  'P_23'  'P_24'  'P_25'  'P_26'  'P_27'  'P_28'  'P_29'  'P_30' 
--------------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- 
       1   2   4   5   6                                                                        
       8   7   9   10   11                                                                        

Answer 2

甲骨文设置：

CREATE TABLE table_name (Family_code, Parent_Family_Code) AS 
SELECT 1, 2 FROM DUAL UNION ALL 
SELECT 2, 4 FROM DUAL UNION ALL 
SELECT 3, 6 FROM DUAL UNION ALL 
SELECT 6, NULL FROM DUAL UNION ALL 
SELECT 4, 3 FROM DUAL UNION ALL 
SELECT 4, 5 FROM DUAL UNION ALL 
SELECT 5, NULL FROM DUAL UNION ALL 
SELECT 8, 7 FROM DUAL UNION ALL 
SELECT 7, 9 FROM DUAL UNION ALL 
SELECT 9, 10 FROM DUAL UNION ALL 
SELECT 10, 11 FROM DUAL UNION ALL 
SELECT 11, NULL FROM DUAL; 

查询：

SELECT TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth, NULL, 1)) AS family_code, 
     CASE WHEN max_depth > 1 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 1, NULL, 1)) END AS p1, 
     CASE WHEN max_depth > 2 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 2, NULL, 1)) END AS p2, 
     CASE WHEN max_depth > 3 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 3, NULL, 1)) END AS p3, 
     CASE WHEN max_depth > 4 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 4, NULL, 1)) END AS p4, 
     CASE WHEN max_depth > 5 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 5, NULL, 1)) END AS p5, 
     CASE WHEN max_depth > 6 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 6, NULL, 1)) END AS p6, 
     CASE WHEN max_depth > 7 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 7, NULL, 1)) END AS p7, 
     CASE WHEN max_depth > 8 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 8, NULL, 1)) END AS p8, 
     CASE WHEN max_depth > 9 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 9, NULL, 1)) END AS p9, 
     CASE WHEN max_depth > 10 THEN TO_NUMBER(REGEXP_SUBSTR(path, '/(\d+)', 1, max_depth - 10, NULL, 1)) END AS p10 
FROM (
    SELECT SYS_CONNECT_BY_PATH(Family_code, '/') AS path, 
     LEVEL AS max_depth 
    FROM table_name 
    WHERE CONNECT_BY_ISLEAF = 1 
    CONNECT BY PRIOR Family_Code = Parent_Family_Code 
    START WITH Parent_Family_Code IS NULL 
); 

输出：

FAMILY_CODE   P1   P2   P3   P4   P5   P6   P7   P8   P9  P10 
----------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- 
      1   2   4   5                    
      1   2   4   3   6                 
      8   7   9   10   11