jra*_*ara 7 sql oracle hierarchy oracle10g common-table-expression
我有这样一张桌子:
+---------+--------+
| EMP_ID | MGR_iD |
+---------+--------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 2 |
| 6 | 2 |
| 7 | 3 |
| 8 | 5 |
| 9 | 7 |
| 10 | 5 |
| 11 | 7 |
| 12 | 9 |
| 13 | 9 |
| 14 | 9 |
+---------+--------+
我正在尝试解析此相邻列表以生成以下结果集:
| EMP_ID | MGR_ID | LV | LEVEL1 | LEVEL2 | LEVEL3 | LEVEL4 | LEVEL5 |
---------------------------------------------------------------------
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 1 | 3 | 3 | 3 | 3 |
| 4 | 2 | 3 | 1 | 2 | 4 | 4 | 4 |
| 5 | 2 | 3 | 1 | 2 | 5 | 5 | 5 |
| 6 | 2 | 3 | 1 | 2 | 6 | 6 | 6 |
| 7 | 3 | 3 | 1 | 3 | 7 | 7 | 7 |
| 8 | 5 | 4 | 1 | 2 | 5 | 8 | 8 |
| 9 | 7 | 4 | 1 | 3 | 7 | 9 | 9 |
| 10 | 5 | 4 | 1 | 2 | 5 | 10 | 10 |
| 11 | 7 | 4 | 1 | 3 | 7 | 11 | 11 |
| 12 | 9 | 5 | 1 | 3 | 7 | 9 | 12 |
| 13 | 9 | 5 | 1 | 3 | 7 | 9 | 13 |
| 14 | 9 | 5 | 1 | 3 | 7 | 9 | 14 |
这是我到目前为止所取得的成就:
create table PC (
EMP_ID NUMBER NULL,
MGR_ID NUMBER NULL
);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (1.0, 1.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (2.0, 1.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (3.0, 1.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (4.0, 2.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (5.0, 2.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (6.0, 2.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (7.0, 3.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (8.0, 5.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (9.0, 7.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (10.0, 5.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (11.0, 7.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (12.0, 9.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (13.0, 9.0);
INSERT INTO PC (EMP_ID, MGR_ID)
VALUES (14.0, 9.0);
和查询:
WITH PERSON_HIER AS
(
SELECT 1 as level1,
CAST(NULL AS NUMBER) as level2,
CAST(NULL AS NUMBER) as level3,
CAST(NULL AS NUMBER) as level4,
CAST(NULL AS NUMBER) as level5
FROM PC
WHERE EMP_ID = 1 AND MGR_ID = 1
UNION ALL
SELECT L1.EMP_ID AS LEVEL1,
L2.EMP_ID AS LEVEL2,
L3.EMP_ID AS LEVEL3,
L4.EMP_ID AS LEVEL4,
L5.EMP_ID AS LEVEL5
FROM PC L1
LEFT OUTER JOIN PC L2 ON (L1.EMP_ID = L2.MGR_ID AND L2.EMP_ID != L1.EMP_ID)
LEFT OUTER JOIN PC L3 ON (L2.EMP_ID = L3.MGR_ID)
LEFT OUTER JOIN PC L4 ON (L3.EMP_ID = L4.MGR_ID)
LEFT OUTER JOIN PC L5 ON (L4.EMP_ID = L5.MGR_ID)
WHERE L1.MGR_ID = L1.EMP_ID
)
SELECT level1,
coalesce(level2, level1) AS LEVEL2,
coalesce(level3, level2, level1) AS LEVEL3,
coalesce(level4, level3, level2, level1) AS LEVEL4,
coalesce(level5, level4, level3, level2, level1) AS LEVEL5
FROM PERSON_HIER
order by level5
我正在使用Oracle 10g,因此Oracle 10g中无法实现递归CTE或Hierarchial查询
您可以使用CONNECT BY使用单个分层查询:
SQL> SELECT root,
2 MAX(DECODE(lvl, max_lvl, mgr_id)) lvl1,
3 MAX(DECODE(lvl, max_lvl, emp_id)) lvl2,
4 MAX(DECODE(lvl, greatest(max_lvl-1, 1), emp_id)) lvl3,
5 MAX(DECODE(lvl, greatest(max_lvl-2, 1), emp_id)) lvl4,
6 MAX(DECODE(lvl, greatest(max_lvl-3, 1), emp_id)) lvl5
7 FROM (SELECT connect_by_root(emp_id) root,
8 emp_id,
9 mgr_id,
10 level lvl,
11 MAX (level)
12 OVER (PARTITION BY connect_by_root(emp_id)) max_lvl
13 FROM pc
14 CONNECT BY NOCYCLE PRIOR mgr_id = emp_id) v
15 GROUP BY root
16 ORDER BY 1;
ROOT LVL1 LVL2 LVL3 LVL4 LVL5
---- ----- ----- ----- ----- -----
1 1 1 1 1 1
2 1 2 2 2 2
3 1 3 3 3 3
4 1 2 4 4 4
5 1 2 5 5 5
6 1 2 6 6 6
7 1 3 7 7 7
8 1 2 5 8 8
9 1 3 7 9 9
10 1 2 5 10 10
11 1 3 7 11 11
12 1 3 7 9 12
13 1 3 7 9 13
14 1 3 7 9 14
现在是改进的版本（显示正确的层次结构）：
select emp_id, mgr_id, lvl, h,
nvl(substr(h,instr(h,'/',1, 2)+1, instr(h, '/',1, 3)- instr(h,'/',1,2)-1), emp_id) as lvl1,
nvl(substr(h,instr(h,'/',1, 3)+1, instr(h, '/',1, 4)- instr(h,'/',1,3)-1), emp_id) as lvl2,
nvl(substr(h,instr(h,'/',1, 4)+1, instr(h, '/',1, 5)- instr(h,'/',1,4)-1), emp_id) as lvl3,
nvl(substr(h,instr(h,'/',1, 5)+1, instr(h, '/',1, 6) -instr(h,'/',1,5)-1), emp_id) as lvl4,
nvl(substr(h,instr(h,'/',1, 6)+1, instr(h, '/',1, 7) -instr(h,'/',1,6)-1), emp_id) as lvl5
from(
select emp_id, mgr_id , level lvl, sys_connect_by_path(mgr_id, '/')||'/' h
from pc
connect by nocycle prior emp_id = mgr_id
start with emp_id = 1
)
order by emp_id;
EMP_ID MGR_ID LVL H LVL1 LVL2 LVL3 LVL4 LVL5
2 1 1 1/1/ 1 2 2 2 2
3 1 1 1/1/ 1 3 3 3 3
4 2 2 2/1/2/ 1 2 4 4 4
5 2 2 2/1/2/ 1 2 5 5 5
6 2 2 2/1/2/ 1 2 6 6 6
7 3 2 3/1/3/ 1 3 7 7 7
8 5 3 5/1/2/5/ 1 2 5 8 8
9 7 3 7/1/3/7/ 1 3 7 9 9
10 5 3 5/1/2/5/ 1 2 5 10 10
11 7 3 7/1/3/7/ 1 3 7 11 11
12 9 4 9/1/3/7/9/ 1 3 7 9 12
13 9 4 9/1/3/7/9/ 1 3 7 9 13
14 9 4 9/1/3/7/9/ 1 3 7 9 14
15 14 5 14/1/3/7/9/14/ 1 3 7 9 14
这是我的第一次尝试：
select emp_id, mgr_id, lvl, h,
nvl(substr(h,instr(h,' ',1, 1), instr(h, ' ',1, 2)- instr(h,' ',1,1)), emp_id) as lvl1,
nvl(substr(h,instr(h,' ',1, 2), instr(h, ' ',1, 3)- instr(h,' ',1,2)), emp_id) as lvl2,
nvl(substr(h,instr(h,' ',1, 3), instr(h, ' ',1, 4)- instr(h,' ',1,3)), emp_id) as lvl3,
nvl(substr(h,instr(h,' ',1, 4), instr(h, ' ',1, 5)- instr(h,' ',1,4)), emp_id) as lvl4,
nvl(substr(h,instr(h,' ',1, 5), instr(h, ' ',1, 6) -instr(h,' ',1,5)), emp_id) as lvl5
from(
select emp_id, mgr_id , level lvl, sys_connect_by_path(mgr_id, ' ') h
from pc
connect by nocycle prior emp_id = mgr_id
start with emp_id = 1
)
order by emp_id;