Re: negative queries puzzle

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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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