DOH!  Answered my own question...

        select max(MYCOUNT)-min(MYCOUNT)
        from MYTABLE
        group by round(rownum/2+.5)*2
        UNION
        select max(MYCOUNT)-min(MYCOUNT)
        from MYTABLE
        group by round((rownum-1)/2+.5)*2

If I had just experimented 1 more minute before writing that message....
<blush>

Rich Jesse                          System/Database Administrator
[EMAIL PROTECTED]             Quad/Tech International, Sussex, WI USA

-----Original Message-----
Sent: Monday, June 18, 2001 14:34
To: '[EMAIL PROTECTED]'


Hi all,

I've got a table, MYTABLE:

        MYCOUNT number
        MYDATE          date, UNIQUE

e.g:
        MYCOUNT MYDATE
        12509           06/17/2001 12:01
        12934           06/17/2001 12:06
        13452           06/17/2001 12:11
        13455           06/17/2001 12:16

I need to get the difference between MYCOUNT on two consecutive rows
(consecutive MYDATEs).  MYCOUNT cannot be less than the MYCOUNT
chronologically before it.

Is there a simple SELECT that I can use for this WITHOUT a cursor?  I've
come close:

        select max(MYCOUNT)-min(MYCOUNT)
        from MYTABLE
        group by round(rownum/2+.5)*2

But this only gives me diffs from rows 1, 2-3, 4-5, 6-7, etc., while I need
the diffs from rows 1-2, 2-3, 3-4, 4-5...  AAAALLLMOST!

TIA,
Rich Jesse                          System/Database Administrator
[EMAIL PROTECTED]             Quad/Tech International, Sussex, WI USA
-- 
Please see the official ORACLE-L FAQ: http://www.orafaq.com
-- 
Author: Jesse, Rich
  INET: [EMAIL PROTECTED]

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