Re: MySQL 5.0: Tricky ranking requirement

[Bill Karwin]

SELECT f0.familynumber, SUM(score) AS totalscore
FROM (SELECT f1.familynumber, f1.name, f1.score, count(*)
FROM f AS f1
LEFT JOIN f AS f2 ON f1.familynumber = f2.familynumber
AND f1.score <= f2.score
GROUP BY f1.familynumber, f1.name
HAVING COUNT(*) <= 3) AS f0
GROUP BY f0.familynumber

Regards,
Bill K.



Edited 2 time(s). Last edit at 06/27/2006 01:56AM by Bill Karwin.