Re: FUNCTION problem
От | Bill Moran |
---|---|
Тема | Re: FUNCTION problem |
Дата | |
Msg-id | 406D64DF.1030608@potentialtech.com обсуждение исходный текст |
Ответ на | FUNCTION problem (Sky <sky@sylvio.hu>) |
Список | pgsql-general |
Sky wrote: > HI everybody ! > > I have a problem, but I don't know the solution: > > CREATE TABLE person( > user_id SERIAL NOT NULL, > uid CHARACTER(20) NOT NULL, > pwd CHARACTER(20) NOT NULL, > PRIMARY KEY (user_id) > ); > > OK, That's right... > > CREATE FUNCTION getuserid (CHARACTER(20),CHARACTER(20)) > RETURNS SETOF INTEGER > AS > ' > SELECT user_id FROM person WHERE uid=$1 AND pwd=$2; > ' > LANGUAGE 'sql'; > > :-( > > ERROR: Unable to identify an operator '=$' for types 'character' and > 'integer > You will have to retype this query using an explicit cast. You don't state what version of Postgres you're using, but I'll bet that it's 7.3 or older. Be a little more liberal with spaces to seperate the tokens in your statement, i.e.: SELECT user_id FROM person WHERE uid = $1 AND pwd = $2; It seems to me that 7.3 and older don't parse quite as intelligently as 7.4 does (which would explain why other people are saying "it works for me") What appears to be happening is that Postgres 7.3 looks at uid=$1 and breaks it down into uid =$ 1, but (unless you created one) it doesn't know anything about how to use =$ as a comparison, so it throws an error. 7.4 seems to get this right more often, but that may be a bug in the other direction ... I mean, what if you defind a =$ operator and really want to compare uid =$ 1? I think the real solution is to write your SQL so it's unambiguious to the parser. I saw this as a suggestion for C programming a few years ago, that you always seperate tokens with space (even if not strictly necessary) to make it unambiguous to the parser, as well as easier for humans to read. I think it's good advice all around. -- Bill Moran Potential Technologies http://www.potentialtech.com
В списке pgsql-general по дате отправления: