1创建这样的表
CREATE TABLE oracle_connet_by_test ( item_id NUMBER PRIMARY KEY, parent_id NUMBER, item_desc VARCHAR2(8)); INSERT INTO oracle_connet_by_test VALUES(1, 0, 'AAA'); INSERT INTO oracle_connet_by_test VALUES(2, 0, 'BBB'); INSERT INTO oracle_connet_by_test VALUES(3, 1, 'CCC'); INSERT INTO oracle_connet_by_test VALUES(4, 2, 'DDD'); INSERT INTO oracle_connet_by_test VALUES(5, 1, 'EEE'); INSERT INTO oracle_connet_by_test VALUES(6, 3, 'FFF'); INSERT INTO oracle_connet_by_test VALUES(7, 3, 'GGG'); commit;
2当我运行此脚本时,输出将是
SQL> SELECT LEVEL, 2 ITEM_ID, 3 PARENT_ID, LPAD(' ', 4 * (LEVEL - 1)) || ITEM_DESC AS ITEM_DESC 4 FROM ORACLE_CONNET_BY_TEST 5 START WITH parent_Id = 0 6 CONNECT BY PRIOR ITEM_ID = PARENT_ID 7 ; LEVEL ITEM_ID PARENT_ID ITEM_DESC ---------- ---------- ---------- ----------- 1 1 0 AAA 2 3 1 CCC 3 6 3 FFF 3 7 3 GGG 2 5 1 EEE 1 2 0 BBB 2 4 2 DDD
3问题:如果我想要这个结果,我该如何编写代码?
LEVEL ITEM_ID PARENT_ID ITEM_DESC SUPER_ID ---------- ---------- ---------- ---------------- ---------- 1 1 0 AAA 1 2 3 1 CCC 1 3 6 3 FFF 1 3 7 3 GGG 1 2 5 1 EEE 1 1 2 0 BBB 2 2 4 2 DDD 2
答案 0 :(得分:6)
使用connect_by_root:
SELECT LEVEL,
ITEM_ID,
PARENT_ID, LPAD(' ', 4 * (LEVEL - 1)) || ITEM_DESC AS ITEM_DESC
, connect_by_root item_id AS SUPER_ID
FROM ORACLE_CONNET_BY_TEST
START WITH parent_Id = 0
CONNECT BY PRIOR ITEM_ID = PARENT_ID
;