Analytic Windowing - Positive/Negative Range

Question

Hi,
In analytic functions, we can say
RANGE BETWEEN  x  PRECEDING
      AND      y  PRECEDINGwhere x and y are expressions (not necessarily literals), as long as x and do not evaluate to negative numbers. If we know when we write the code that x and/or y is negative, ithen we can use FOLLOWING instead of PRECEDING, and if we don't know until run-time whether x or y will be negative, we can compute the function both ways, and use a CASE expression to choose one or the other. I was just wondering if there was a more elegant way do RANGE BETWEEN when x and/or y might be negative.
Here's a specific example:
Yesterday, there was a question {message:id=10868410} about how to find how many children (immediate children, not descendants 2 or more levels down) each node in a tree has. OP specifically asked for an analytic solution. I don't believe analytic functions are the best solution to this particular problem, but I've been trying to find an analytic solution anyway, just for curiosity.
So, again, the problem is, using the standard scott.emp table (where empno is a unique postive number, and NULL mgr indicates no parent), is there a purely analytic query that can get the same results as
SELECT       ename           -- Query 1: no analytics
,       empno
,       mgr
          SELECT  COUNT (*)
          FROM     scott.emp
          WHERE     mgr     = m.empno
       )          AS children
FROM       scott.emp  m
ORDER BY  mgr          NULLS FIRST
,            empno
;Output:
ENAME           EMPNO        MGR   CHILDREN
KING             7839                     3
SCOTT            7788       7566          1
FORD             7902       7566          1
ALLEN            7499       7698          0
WARD             7521       7698          0
MARTIN           7654       7698          0
TURNER           7844       7698          0
JAMES            7900       7698          0
MILLER           7934       7782          0
ADAMS            7876       7788          0
JONES            7566       7839          2
BLAKE            7698       7839          5
CLARK            7782       7839          1
SMITH            7369       7902          0I would love to do something like this:
SELECT       ename          -- Query 2: ***** WRONG ! *****  raises ORA-01428
,       empno
,       mgr
,       COUNT (*) OVER ( ORDER BY         mgr
                  RANGE BETWEEN  empno - NVL (mgr, 0)  PRECEDING
                        AND       empno - NVL (mgr, 0)  PRECEDING
                )     AS children
FROM      scott.emp
ORDER BY  mgr, empno
;but negative RANGEs aren't allowed, so the best I could do using this approach was
SELECT       ename          -- Query 3
,       empno
,       mgr
,       CASE
           WHEN empno < mgr
           THEN COUNT (*) OVER ( ORDER BY      NVL (mgr, 0)
                            RANGE BETWEEN ABS (empno - NVL (mgr, 0))  PRECEDING
                                     AND     ABS (empno - NVL (mgr, 0))  PRECEDING
           ELSE COUNT (*) OVER ( ORDER BY         NVL (mgr, 0)
                             RANGE BETWEEN ABS (empno - NVL (mgr, 0))  FOLLOWING
                                  AND       ABS (empno - NVL (mgr, 0))  FOLLOWING
          END     AS children
FROM      scott.emp
ORDER BY  NVL (mgr, 0), empno
;Is there a more efficient and/or more elegant way to handle possibly negative ranges?
Can you think of other analytic solutions to this particular problem?
OP in {message:id=10868481} had one idea, that involved a CONNECT BY query.
I posted a soltuion using the aggregate COUNT function. Any query that uses an aggregate function (such as COUNT) which has an analytic twin can be re-written to use the analytic twin, and SELECT DISTINCT instead of GROUP BY.
Any other ideas?
I'm interested in any version of Oracle.
Thanks.

Answer 1

Hi,
I don't have a database at the moment but to solve the preceding or folowing problem, I think you can use one of the two always, and just sort ascending or descending as applicable.
So some thing like: (not tested)
SELECT       ename          -- Query 3
,       empno
,       mgr
,      COUNT (*) OVER ( ORDER BY      case when empno < mgr then NVL (mgr, 0) else -NVL (mgr, 0) end
                            RANGE BETWEEN ABS (empno - NVL (mgr, 0))  PRECEDING
                                     AND     ABS (empno - NVL (mgr, 0))  PRECEDING
     AS children
FROM      scott.emp
ORDER BY  NVL (mgr, 0), empno
;Regards,
Peter