具有2个表的递归CTE在Oracle / SQL中不起作用

Question

There are 2 tables Department and subdepartment which have id in common. I am trying to recursively fetch all the ids reporting to AB directly and indirectly. BC is reporting to AB, hence 4,5,6 are indirectly reporting to AB, likewise fetching till the last id.

Tried the below recursive CTE query but I am getting the result of only the first level. Seems recursive query is not executing.

I am not sure what is wrong in the query. Can someone help me in spotting the error.

Department

Name     id
AB          1
AB          2
AB          3
BC          4
BC          5
BC          6
CD          7
CD          8
EF          9
EF         10
EF         11

Subdepartment

ID      Reporting
1
2
3         BC
4
5         CD
6
7
8         EF
9
10
11

Query:

With reportinghierarchy (Name, Id, Reporting, Level) As
(

--Anchor
Select A.name,A.id,reporting,0 from department A, subdepartment B
where A.id=B.id and A.name='AB'

Union All

--Recursive member
Select C.name,C.id,D.reporting, Level+1 from department C, subdepartment D
Inner Join  reportinghierarchy R
On (C.Name = R.reporting)
Where C.name != 'AB' and C.Id =D.id
And R.Reporting is not null
)
Select * from reportinghierarchy

Current Output :

Name Id  Reporting  Level
AB   1                0
AB   2                0
AB   3   BC           0

Expected output :

Name     id     Reporting  Level
AB        1                  0
AB        2                  0
AB        3       BC         0
BC        4                  1
BC        5       CD         1
BC        6                  1
CD        7                  2
CD        8       EF         2
EF        9                  3
EF       10                  3
EF       11                  3

Answer 1

Hmmm, "horrible data structure" comes to mind. This approach gets one row per "reporting" name to use for the recursive CTE portion. It then joins the level back to the original data.

with ds as (
      select d.name, d.id, sd.reporting
      from department d join
           subdepartment sd
           on d.id = sd.id
     ),
     nd as (
      select d.name, sd.reporting
      from ds
      where sd.reporting is not null
     ),
     cte as (
      select ds.name, nd.reporting, 0 as lev
      from nd
      where not exists (select 1 from nd nd2 where nd2.reporting = nd.name)
      union all
      select nd.name, nd.reporting, lev + 1
      from cte join
           nd
           on nd.name = cte.reporting 
    )
select ds.*, cte.lev
from ds join
     cte
     on ds.name = cte.name;

Also, learn to use proper, explicit JOIN syntax. It has been the standard syntax for decades.

Answer 2

Your original query was actually VERY, VERY close to working. The reasons it didn't work are:

1. You used the keyword LEVEL as a column name without quoting it. In Oracle LEVEL has specific meaning, and using it out of context causes the parser no end of headaches. I've changed it to LVL , which works fine.
2. In the recursive half of the UNION you mixed old-style and new-style joins . This is a huge problem and should never be done. Either use "old-style" implied joins, or use "new-style" explicit joins. To keep as close to your original as possible I used implicit joins, but good coding practice says you should use explicit joins all the time.

The corrected query is:

With reportinghierarchy (Name, Id, Reporting, lvl) As
(

--Anchor
Select A.name,A.id,reporting,0 from department A, subdepartment B
where A.id=B.id and A.name='AB'

Union All

--Recursive member
Select C.name,C.id,D.reporting, lvl+1 from department C, subdepartment D, reportinghierarchy R
Where C.name != 'AB' and C.Id =D.id and C.Name = R.reporting
And R.Reporting is not null
)
Select * from reportinghierarchy;

Given the above, the following results are returned, which appear to match your desired results:

NAME    ID  REPORTING   LVL
AB      1   (null)      0
AB      2   (null)      0
AB      3   BC          0
BC      4   (null)      1
BC      5   CD          1
BC      6   (null)      1
CD      7   (null)      2
CD      8   EF          2
EF      9   (null)      3
EF      10  (null)      3
EF      11  (null)      3

SQLFiddle here

Best of luck.