* 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/>

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