我有一张这样的桌子:

+---------+--------+
| 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产生以下结果集:

| 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 |


这是到目前为止我设法做到的:

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


我使用的是Oracle 10g,因此在Oracle 10g中无法进行递归CTE或层次结构查询

最佳答案

现在是一个改进的版本(显示正确的层级):

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


SQL FIDDLE

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;


See SQLFiddle here

08-05 04:58