At 05:38 PM 9/26/2008, Oliveiros Cristina wrote:
In-Reply-To: <[EMAIL PROTECTED]>
References: <[EMAIL PROTECTED]>
<[EMAIL PROTECTED]>
<[EMAIL PROTECTED]>
Howdy, Steve.
SELECT id
FROM dummy a
NATURAL JOIN (
SELECT fkey_id,name
FROM dummy
GROUP BY fkey_id,name
HAVING COUNT(*) > 1 AND SUM(id) = (MAX(id) + MIN(id)) * (MAX(id) -
MIN(id) + 1) / 2
) b
ORDER BY id;
In your table you just have duplicates? Or you may have triplicates?
And quadruplicates? And in general n-uplicates? At the time, I thought
you might have n-uplicates, so I designed the query to be as general
as possible to handle all that cases, from which duplicates are a
particular case, but now i am wondering if you don't have more than
duplicates.
In my specific case it turns out I only had duplicates, but there could
have been n-plicates, so your code is still correct for my use-case
(though I didn't say that in my OP).
Well, anyway the idea is as follows
The sum of a sequence is given by first + last / 2 * n, with n = last
- first + 1, OK ?
I *love* your application of that formula. It's rare for me to be able
to use "real" math in SQL, so this was a pleasure to read (and
understand!)
Thanks again to Richard and Oliveiros for a truly educating experience!
I hope some others were similarly enlightened.
With gratitude,
Steve
--
Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
To make changes to your subscription:
http://www.postgresql.org/mailpref/pgsql-sql
