我的数据如下：
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
这是分层数据，我将查询根复合密钥（在本例中为09/10 10000 A1234）; FLAG字段指的是LKEYx键标识的“对象”。可以有任意数量的嵌套级别。 （请注意，只要保留层次结构，KEY1和KEY3字段就不必是不变的，如上例所示。）
我想要检索的是叶子节点，但，如果叶子的父KEY2长度与LKEY2相同或包含X第二个字符，然后返回直接父母。在这种情况下，我们还需要将记录标记为可选...所以，像这样：
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
我已经编写了一个执行此操作的查询，但它并不漂亮。此外，它假设所有叶节点在树下面处于同一级别，以便区分可选记录;但是，这不一定是真的。我的查询如下：
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');
queryKeys的原因是因为它在较大的查询中的其他位置使用，并且可以包含多个记录。 maxTree部分是问题，超出了它的一般怪癖！
现在，这篇文章标题的原因是因为如果我可以引用父级的FLAG字段，这个查询可以更简单地很多。我尝试了JOIN这个想法的方法 - 将树与自己连接在相关的键上 - 但是，除非我弄错了，否则会导致递归问题，你不得不迭代树以查找正确的父键（因为KEYx和LKEYx字段都定义了记录的完整复合键。）
（P.S。使用Oracle 10gR2，如果它有所作为。）
答案 0 :(得分：3)
只需使用：
PRIOR FLAG
它会准确地给你你想要的东西 - 父行的标志字段。
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
(...)
答案 1 :(得分：2)
我认为您的帖子归结为“我如何在分层查询中引用父行的FLAG属性？”
我不知道我提出的SQL是否正确。如果不是，我道歉。但总的来说，这是我的方法：
在层次结构的每个级别，我将所有键（SYS_CONNECT_BY_PATH）串在一起。然后使用SUBSTR，INSTR和LEVEL，我SUBSTR将相当于父级关键字的内容删除。最后，在PARENT_FLAG的定义中，我选择了一行的FLAG，其中的键与此SUBSTR渐出键匹配。
设定：
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');
查询：
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
;
结果：
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>
正如我所说，我并非100％确定我完全了解您的数据模型，但我希望您能够遵循我的方法。