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;
The GROUP BY clause is to associate records that have the same fkey_id and name The COUNT(*) > 1 eliminates the situations when there is just one. Now, about the equality, now i am thinking and maybe it is a bazooka to kill a fly. :) 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. 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 ? So, if the set of ids is sequencial, its sum must equal that expression. It's basically that. But I am now wondering now that I might have misunderstood what your requests were... If you just have duplicates, then maybe it is cleaner to substitute that clause by something simpler, like MAX(id) - MIN(id) = 1 I dunno if I fully answered your questions, but if I didn't feel free to ask Best, Oliveiros > > -- We are going to have peace even if we have to fight for it. - General Dwight D. Eisenhower Teremos paz, nem que tenhamos de lutar por ela - General Dwight D. Eisenhower
