Complex hierarchy chain.

Question

Hi All,
This is reference to the thread Hierarchy order number. I am using Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production. Frank helped to obtain the hierrachy relationship between the parts.
During QA testing they found the below scenario and it is called 'Complex hierrarchy order'.
Here is the sample data creation.
CREATE TABLE BL_SAMPLE_SS_DATA
AS
SELECT 'A' AS OLD_PART_NO,
       '2' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT 'B' AS OLD_PART_NO,
       '1' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT 'B' AS OLD_PART_NO,
       '2' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT 'C' AS OLD_PART_NO,
       '2' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT '2' AS OLD_PART_NO,
       'D' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT '2' AS OLD_PART_NO,
       'E' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT 'D' AS OLD_PART_NO,
       '3' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT 'E' AS OLD_PART_NO,
       '4' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT '4' AS OLD_PART_NO,
       '5' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT '5' AS OLD_PART_NO,
       '6' AS NEW_PART_NO
  FROM DUAL
UNION
SELECT '5' AS OLD_PART_NO,
       '7' AS NEW_PART_NO
  FROM DUAL;
Explanation about the above sample data:
1. The part 'A' was first manufactured later the part A is replaced by part number 2.
2. The part B is replaced by two part number 1 and 2. ( The customer can pick either 1 or 2)
    it means Part B is replaced by two parts 1 and 2. for example if Part B is a car front mirror passenger wiper with 14 inch size, after sometime Ford released the same car front mirror passnger wiper in 2 sizes one is part 1 (8 inch) and next one is part 2 (14 inch). we will show this description of part in part details page. so we should show both like part B is replaced with 1 and 2 in chain. Based on customer need, they will pick one.
3. Part C is replaced by 2.
4. Part number 2 is replaced by two parts D and E (same as #2).
5. Part D is replaced by 3 and Part E is replaced by 4.
6. part 4 is replaced by 5 and Part 5 is replaced by 6 and 7.
so the hierrachy would be like below
A -> 2
B -> 1 and 2
C -> 2
2 -> D and E
D -> 3
E -> 4
4 -> 5
5 -> 6 and 7
so the output should contain data like below for each present in the chain. Below output table contain chain for random two parts like this i need to show for all the parts present in that chain.
PARTNO
OLD PART NO
NEW PART NO
B
A
2
B
B
1
B
B
2
B
C
2
B
2
D
B
2
E
B
D
3
B
E
4
B
4
5
B
5
6
B
5
7
5
A
2
5
B
1
5
B
2
5
C
2
5
2
D
5
2
E
5
D
3
5
E
4
5
4
5
5
5
6
5
5
7
Thanks All
Anyone please guide me to do this. Thanks All

Answer 1

Really sorry about not mention in initial post. The SEQ_NO should be assigned based on hierarchy order of the parts. For example,  the hierarchy of above sample data like below.
OLD PART
  NO
NEW PART NO
A
2
B
1
B
2
C
2
2
D
2
E
D
3
E
4
4
5
5
6
5
7
So regardless of the PART_NO we should show in the table in above order. The SEQ_NO will be used when application render the data from table like order by seq number. so that the hierarchy wont change.
for an example and keep it short i will take 2 part numbers B and 5. The data in target table (PARTS) will be like below
PARTNO
OLD PART NO
NEW PART NO
SEQ_NO
B
A
2
1
B
B
1
2
B
B
2
3
B
C
2
4
B
2
D
5
B
2
E
6
B
D
3
7
B
E
4
8
B
4
5
9
B
5
6
10
B
5
7
11
5
A
2
51
5
B
1
52
5
B
2
53
5
C
2
54
5
2
D
55
5
2
E
56
5
D
3
57
5
E
4
58
5
4
5
59
5
5
6
60
5
5
7
61
WHen application does "SELECT * FROM PARTS WHERE PART_NO = 5 ORDER BY SEQ_NO" .The result will show as per the order in case if the order in the target table is wrong. Hope this helps.