Yes, for each (S, I) pair the goal is to efficiently find the next largest integer associated with S in T. For the highest integer I associated with S in T, there is no next larger.
Thanks, Mike Spreitzer Peter Brawley <peter.braw...@earthlink.net> 06/20/09 08:56 AM Please respond to peter.braw...@earthlink.net To Mike Spreitzer/Watson/i...@ibmus cc mysql@lists.mysql.com Subject Re: how to efficiently query for the next in MySQL Community Edition 5.1.34? Mike >J holding the next integer that T has for S You mean for each i, the next value of i with that s? >(U having no row for the last integer of each string). I do not understand that at all. PB Mike Spreitzer wrote: Suppose I have a table T with two column, S holding strings (say, VARCHAR(200)) and I holding integers. No row appears twice. A given string appears many times, on average about 100 times. Suppose I have millions of rows. I want to make a table U holding those same columns plus one more, J holding the next integer that T has for S (U having no row for the last integer of each string). I could index T on (S,I) and write this query as select t1.*, t2.I as J from T as t1, T as t2 where t1.S=t2.S and t1.I < t2.I and not exists (select * from T as t12 where t12.S=t1.S and t1.I < t12.I and t12.I < t2.I) but the query planner says this is quite expensive to run: it will enumerate all of T as t1, do a nested enumeration of all t2's entries for S=t1.S, and inside that do a further nested enumeration of t12's entries for S=t1.S --- costing about 10,000 times the size of T. There has to be a better way! Thanks, Mike Spreitzer No virus found in this incoming message. Checked by AVG - www.avg.com Version: 8.5.364 / Virus Database: 270.12.80/2187 - Release Date: 06/19/09 06:53:00
