* A. Pagaltzis <[EMAIL PROTECTED]> [2006-05-03 00:30]: > I tried to do it with a join to see if that would work better, > but I’m too frazzled to figure it out right now.
I must be more frazzled than I thought.
SELECT
n1.name,
COUNT( n2.name ) rank
FROM names n1
CROSS JOIN names n2
WHERE n2.name < n1.name
GROUP BY n1.name
ORDER BY rank;
Regards,
--
Aristotle Pagaltzis // <http://plasmasturm.org/>

