Re: 8.0.3 regexp_replace()...
От | Stephan Szabo |
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Тема | Re: 8.0.3 regexp_replace()... |
Дата | |
Msg-id | 20060131100235.I54974@megazone.bigpanda.com обсуждение исходный текст |
Ответ на | Re: 8.0.3 regexp_replace()... ("rlee0001" <robeddielee@hotmail.com>) |
Список | pgsql-general |
On Mon, 30 Jan 2006, rlee0001 wrote: > I did get the code working. The function DDL follows: > > CREATE OR REPLACE FUNCTION "webadmin"."regexp_replacex" (source > varchar, pattern varchar, replacement varchar) RETURNS varchar AS > $body$ > DECLARE > retvalue VARCHAR; > BEGIN > retvalue = "source"; > LOOP > retvalue = REPLACE(retvalue, COALESCE(SUBSTRING(retvalue FROM > "pattern"), ''), "replacement"); > EXIT WHEN retvalue = REPLACE(retvalue, COALESCE(SUBSTRING(retvalue > FROM "pattern"), ''), "replacement"); > END LOOP; > RETURN retvalue; > END; > $body$ > LANGUAGE 'plpgsql' VOLATILE CALLED ON NULL INPUT SECURITY INVOKER; > > The problem was that SUBSTRING returns NULL if it cannot find any > matches for the pattern and when the second parameter to REPLACE > returns NULL, REPLACE returns NULL (which is idiotic). Why do you say that? I'd say that's precisely the most reasonable answer. You're asking to replace an unknown portion of a string (since NULL is unknown, it might match some portion of the string, it might not) with something else. The answer to that seems pretty unknown to me.
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