问题
I've been struggling a bit about how to deal with this situation:
I have a table structured as follow:
Family_code | Parent_Family_Code | ....
1 2
2 4
3 6
4 3
......................
When a user is searching for a specific family code, I need to return the entire path (up to 10 levels max) , so for example for family_code = 1 I'll need:
Family_code | parent_1 | p_2 | p_3 | p_4 | p_5 | .....
1 2 4 3 6 null null.....
I know I can use sys_connect_by_path() which will bring me the expected result but as a string, and not as separate columns which is something I'll prefer to avoid.
This can also be done with 10 left joins to the same table, or the use of LEAD()/LAG() functions which will include a lot of sub queries and will make a messy and unreadable query, but then again, this will be more heavy then it should be and I need to simplify it as I can.
I've come up with a solution using substr() function(the length of the codes will always be 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
But I would like a solution without the use of substrings since it will be harder to do any maintaince on it when other developers will touch it . So basically my question is - how do I select the entire path as different columns without the use of substrings?
回答1:
A slightly modified query from @MT0 answer, using PIVOT clause.
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 */
)
);
A result of the query for sample data from @MT0 answer (@MT0, thank you for providing sample data):
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
回答2:
Oracle Setup:
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;
Query:
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
);
Output:
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
来源：https://stackoverflow.com/questions/36964508/how-to-get-the-path-of-an-hierarchy-table