Re: negative queries puzzle
От | Stephan Szabo |
---|---|
Тема | Re: negative queries puzzle |
Дата | |
Msg-id | 20020731130557.A19958-100000@megazone23.bigpanda.com обсуждение исходный текст |
Ответ на | negative queries puzzle (Jinn Koriech <lists@idealint.co.uk>) |
Список | pgsql-sql |
On 31 Jul 2002, Jinn Koriech wrote: > hi all, > > here's a query i've never been able to improve: > > i have an old data set and a new data set - in this case uk postcodes > with eastings and northings. i want to extract the new and changed > postcodes from the new set. to get the changed entries i use a join and > it works okay: > > SELECT n.postcode, n.easting, n.northing FROM v_postcode_new n, > v_postcode_old o WHERE n.postcode = o.postcode AND (n.easting <> > o.lattitude OR n.northing <> o.longitude); > > > but then to get the entirely new items out i use a sub query which takes > for ever > > SELECT DISTINCT * FROM v_postcode_new WHERE postcode NOT IN ( SELECT > postcode FROM v_postcode_old ) ORDER BY postcode ASC; > > does anyone know of a quicker way to accomplish this? i guess there > must be some cleaver way around it, but it's beyond me. Hmm, a couple of possible other queries: -- Do you really need the distinct? select distinct * from v_postcode_new where not exists (select * from v_postcode_old where v_postcode_old.postcode= v_postcode_new.postcode); Or maybe (just thought of this, think it should work, but am not entirely sure) select distinct v_postcode_new.* fromv_postcode_new left outer join v_postcode_old using(postcode)where v_postcode_old.postcodeis null;
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