树形查询新特性CONNECT_BY_ISLEAF的9i实现方式

 在10g中Oracle提供了新的伪列：CONNECT_BY_ISLEAF，通过这个伪列，可以判断当前的记录是否是树的叶节点。
这里描述一下在9i中如何实现相应的功能.
首先构造一个例子：

 

CREATE TABLE T_TREE (ID NUMBER PRIMARY KEY, FATHER_ID NUMBER, NAME VARCHAR2(30));
INSERT INTO T_TREE VALUES (1, 0, 'A');
INSERT INTO T_TREE VALUES (2, 1, 'BC');
INSERT INTO T_TREE VALUES (3, 1, 'DE');
INSERT INTO T_TREE VALUES (4, 1, 'FG');
INSERT INTO T_TREE VALUES (5, 2, 'HIJ');
INSERT INTO T_TREE VALUES (6, 4, 'KLM');
INSERT INTO T_TREE VALUES (7, 6, 'NOPQ');
COMMIT;

 
 
select * from t_tree;
 

4000

	ID	FATHER_ID	NAME
1	1	0	A
2	2	1	BC
3	3	1	DE
4	4	1	FG
5	5	2	HIJ
6	6	4	KLM
7	7	6	NOPQ

 
 
看看CONNECT_BY_ISLEAF的功能：
 

SELECT ID, FATHER_ID, NAME, CONNECT_BY_ISLEAF LEAF
FROM T_TREE
START WITH FATHER_ID = 0
CONNECT BY PRIOR ID = FATHER_ID;

 

	ID	FATHER_ID	NAME	LEAF
1	1	0	A	0
2	2	1	BC	0
3	5	2	HIJ	1
4	3	1	DE	1
5	4	1	FG	0
6	6	4	KLM	0
7	7	6	NOPQ	1

 

SELECT ID, FATHER_ID, NAME, CONNECT_BY_ISLEAF LEAF
FROM T_TREE
START WITH ID = 7
ONNECT BY PRIOR FATHER_ID = ID;

	ID	FATHER_ID	NAME	LEAF
1	7	6	NOPQ	0
2	6	4	KLM	0
3	4	1	FG	0
4	1	0	A	1

 
CONNECT_BY_ISLEAF可以判断当前记录是否是树的叶节点。而这个功能在9i中没有简单的方法来实现，只能通过分析函数来进行判断：
 

SELECT
ID,
FATHER_ID,
NAME,
CASE WHEN LEAD(LEVELS) OVER(ORDER BY RN) > LEVELS THEN 0 ELSE 1 END LEAF
FROM
(
SELECT ROWNUM RN, ID, FATHER_ID, NAME, LEVEL LEVELS
FROM T_TREE
START WITH FATHER_ID = 0
CONNECT BY PRIOR ID = FATHER_ID
);

 

	ID	FATHER_ID	NAME	LEAF
1	1	0	A	0
2	2	1	BC	0
3	5	2	HIJ	1
4	3	1	DE	1
5	4	1	FG	0
6	6	4	KLM	0
7	7	6	NOPQ	1

 

SELECT
ID,
FATHER_ID,
NAME,
CASE WHEN LEAD(LEVELS) OVER(ORDER BY RN) > LEVELS THEN 0 ELSE 1 END LEAF
FROM
(
SELECT ROWNUM RN, ID, FATHER_ID, NAME, LEVEL LEVELS
FROM T_TREE
START WITH ID = 7
CONNECT BY PRIOR FATHER_ID = ID
);

 

	ID	FATHER_ID	NAME	LEAF
1	7	6	NOPQ	0
2	6	4	KLM	0
3	4	1	FG	0
4	1	0	A	1

 利用分析函数可以相对简单的在9i实现CONNECT_BY_ISLEAF伪列的功能。