Reference to parent columns in Oracle CONNECT BY hierarchical query
https://stackoverflow.com/questions/10387124
-
04-06-2021 - |
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I have data that looks like this:
KEY1 KEY2 KEY3 LKEY1 LKEY2 LKEY3 FLAG
====== ========= ====== ====== ========= ====== =====
09/10 10000 A1234 09/10 AU000123 A1234 1
09/10 10000 A1234 09/10 AU000456 A1234 1
09/10 10000 A1234 09/10 AX000001 A1234 1
09/10 AX000001 A1234 09/10 AE000010 A1234 0
09/10 AX000001 A1234 09/10 AE000020 A1234 0
09/10 AX000001 A1234 09/10 AE000030 A1234 0
09/10 10000 A1234 09/10 AX000002 A1234 0
09/10 AX000002 A1234 09/10 AE000040 A1234 0
09/10 10000 A1234 09/10 AU000789 A1234 0
This is hierarchical data, whereby I would be querying against the root composite key (in this case 09/10 10000 A1234); the FLAG field refers to the 'object' identified by the LKEYx keys. There can be any number of levels of nesting. (Note that the KEY1 and KEY3 fields needn't be invariant, as in the example above, as long as the hierarchy is preserved.)
What I want to retrieve are the leaf nodes, but if a leaf's parent KEY2 is the same length as LKEY2 or contains an X as the second character, then return the immediate parent. In this case, we also need to mark the record as optional... So, something like this:
KEY1 KEY2 KEY3 OPTION FLAG
====== ========= ====== ======= =====
09/10 AU000123 A1234 0 1
09/10 AU000456 A1234 0 1
09/10 AX000001 A1234 1 1
09/10 AX000002 A1234 1 0
09/10 AU000789 A1234 0 0
I have written a query that does this, but it's not pretty. Moreover, it makes the assumption that all leaf nodes are at the same level down the tree in order to distinguish the optional records; however, this is not necessarily true. My query is as follows:
with queryKeys as (
select '09/10' key1,
'10000' key2,
'A1234' key3,
from dual
),
subTree as (
select tree.key1,
tree.key2,
tree.key3,
tree.lkey1,
tree.lkey2,
tree.lkey3,
tree.flag,
connect_by_isleaf isLeaf,
level thisLevel
from tree,
queryKeys
start with tree.key1 = queryKeys.key1
and tree.key2 = queryKeys.key2
and tree.key3 = queryKeys.key3
connect by tree.key1 = prior tree.lkey1
and tree.key2 = prior tree.lkey2
and tree.key3 = prior tree.lkey3
),
maxTree as (
select max(thisLevel) maxLevel
from subTree
)
select lkey1 key1,
lkey2 key2,
lkey3 key3,
1 - isLeaf option,
flag
from subTree,
maxTree
where (isLeaf = 1 or thisLevel = maxLevel - 1)
and (length(key2) != length(lkey2) or substr(lkey2, 2, 1) != 'X');
The reason for queryKeys is because it's used elsewhere in a larger query and can contain more than one record. The maxTree part is the problem, beyond its general quirkiness!
Now, the reason for the title of this post is because this query could be made a lot more straightforward if I could refer to the parent's FLAG field. I tried a JOIN approach to this idea -- joining the tree with itself on the relevant keys -- but, unless I'm mistaken, that would result in a recursive problem where you'd keep having to iterate up the tree to find the correct parent keys (as both the KEYx and LKEYx fields define the complete composite key for the record).
(P.S. Using Oracle 10gR2, if it makes a difference.)
Solution
Just use:
PRIOR FLAG
it will give you exactly what you want - the flag field of the parent row.
subTree as (
select tree.key1,
tree.key2,
tree.key3,
tree.lkey1,
tree.lkey2,
tree.lkey3,
tree.flag,
PRIOR TREE.FLAG PRIOR_FLAG
connect_by_isleaf isLeaf,
level thisLevel
from tree,
queryKeys
(...)
OTHER TIPS
I've assumed your post boils down to the question, "How do I reference the FLAG attribute of the parent row in a hierarchical query?"
I don't know the SQL I've come up with is correct. I apologize if it isn't. In general, though, here was my approach:
At each level in the hierarchy, I stringed together all the keys (SYS_CONNECT_BY_PATH). Then using SUBSTR, INSTR and LEVEL, I SUBSTRinged out what amounts to the parent level key. Finally, in the definition of PARENT_FLAG, I select the FLAG of a row whose key matches this SUBSTRinged-out key.
Setup:
CREATE TABLE tree (
key1 VARCHAR2(5)
, key2 VARCHAR2(10)
, key3 VARCHAR2(5)
, lkey1 VARCHAR2(5)
, lkey2 VARCHAR2(10)
, lkey3 VARCHAR2(5)
, flag VARCHAR2(1)
);
INSERT INTO tree VALUES ('09/10','10000','A1234','09/10','AU000123','A1234','1');
INSERT INTO tree VALUES ('09/10','10000','A1234','09/10','AU000456','A1234','1');
INSERT INTO tree VALUES ('09/10','10000','A1234','09/10','AX000001','A1234','1');
INSERT INTO tree VALUES ('09/10','AX000001','A1234','09/10','AE000010','A1234','0');
INSERT INTO tree VALUES ('09/10','AX000001','A1234','09/10','AE000020','A1234','0');
INSERT INTO tree VALUES ('09/10','AX000001','A1234','09/10','AE000030','A1234','0');
INSERT INTO tree VALUES ('09/10','10000','A1234','09/10','AX000002','A1234','0');
INSERT INTO tree VALUES ('09/10','AX000002','A1234','09/10','AE000040','A1234','0');
INSERT INTO tree VALUES ('09/10','10000','A1234','09/10','AU000789','A1234','0');
Query:
COL flag FOR A4
COL same_length FOR A11
COL has_x_2nd FOR A9
COL full_key_path FOR A50
COL parent_key FOR A30
COL parent_flag FOR A11
WITH querykeys AS (
SELECT '09/10' key1
, '10000' key2
, 'A1234' key3
FROM DUAL
)
, subtree1 AS (
SELECT tree.key1
, tree.key2
, tree.key3
, tree.lkey1
, tree.lkey2
, tree.lkey3
, tree.flag
, CONNECT_BY_ISLEAF isleaf
, LEVEL thislevel
, DECODE(LENGTH(tree.key2)
, LENGTH(tree.lkey2), '1'
, '0') same_length
, DECODE(UPPER(SUBSTR(tree.key2,2,1))
, 'X', '1'
, '0') has_x_2nd
, SYS_CONNECT_BY_PATH(tree.key1 || '|' || tree.key2 || '|' || tree.key3,'\')
|| '\'
|| tree.lkey1 || '|' || tree.lkey2 || '|' || tree.lkey3 || '\' full_key_path
FROM tree
, querykeys
START WITH tree.key1 = querykeys.key1
AND tree.key2 = querykeys.key2
AND tree.key3 = querykeys.key3
CONNECT BY tree.key1 = PRIOR tree.lkey1
AND tree.key2 = PRIOR tree.lkey2
AND tree.key3 = PRIOR tree.lkey3
)
, subtree2 AS (
SELECT st1.key1
, st1.key2
, st1.key3
, st1.lkey1
, st1.lkey2
, st1.lkey3
, st1.flag
, st1.isleaf
, st1.thislevel
, st1.same_length
, st1.has_x_2nd
, st1.full_key_path
, SUBSTR(st1.full_key_path
, INSTR(st1.full_key_path,'\',1,st1.thislevel) + 1
, INSTR(st1.full_key_path,'\',1,st1.thislevel + 1)
- INSTR(st1.full_key_path,'\',1,st1.thislevel) - 1) parent_key
FROM subtree1 st1
)
SELECT st2.key1
, st2.key2
, st2.key3
, st2.lkey1
, st2.lkey2
, st2.lkey3
, st2.flag
, st2.isleaf
, st2.thislevel
, st2.same_length
, st2.has_x_2nd
, (SELECT t_prime.flag
FROM tree t_prime
WHERE t_prime.key1 = SUBSTR(st2.parent_key
, 1
, INSTR(st2.parent_key,'|',1,1) - 1)
AND t_prime.key2 = SUBSTR(st2.parent_key
, INSTR(st2.parent_key,'|',1,1) + 1
, INSTR(st2.parent_key,'|',1,2)
- INSTR(st2.parent_key,'|',1,1) - 1)
AND t_prime.key3 = SUBSTR(st2.parent_key
, INSTR(st2.parent_key,'|',1,2) + 1)
-- Following assumes all rows with parent keys have same flag value.
-- Avoids ORA-01427: single-row subquery returns more than one row.
AND ROWNUM = 1) parent_flag
FROM subtree2 st2
;
Result:
KEY1 KEY2 KEY3 LKEY1 LKEY2 LKEY3 FLAG ISLEAF THISLEVEL SAME_LENGTH HAS_X_2ND PARENT_FLAG
----- ---------- ----- ----- ---------- ----- ---- ---------- ---------- ----------- --------- -----------
09/10 10000 A1234 09/10 AU000123 A1234 1 1 1 0 0 1
09/10 10000 A1234 09/10 AU000456 A1234 1 1 1 0 0 1
09/10 10000 A1234 09/10 AU000789 A1234 0 1 1 0 0 1
09/10 10000 A1234 09/10 AX000001 A1234 1 0 1 0 0 1
09/10 AX000001 A1234 09/10 AE000010 A1234 0 1 2 1 1 0
09/10 AX000001 A1234 09/10 AE000020 A1234 0 1 2 1 1 0
09/10 AX000001 A1234 09/10 AE000030 A1234 0 1 2 1 1 0
09/10 10000 A1234 09/10 AX000002 A1234 0 0 1 0 0 1
09/10 AX000002 A1234 09/10 AE000040 A1234 0 1 2 1 1 0
9 rows selected.
SQL>
As I said, I'm not 100% sure I grokked your data model completely, but I hope you're able to follow my approach.
