If you are on 8.1.6 or above, check out the new LEAD sql function.
Ron.
Rich.Jesse@qt
iworld.com To: [EMAIL PROTECTED]
Sent by: cc:
root@fatcity. Subject: RE: Calcs between two rows of
same table. (NEVERMIND)
com
06/18/01
01:45 PM
Please
respond to
ORACLE-L
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
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Author: Jesse, Rich
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