Re: 8.2/8.3 incompatibility

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От Harald Fuchs
Тема Re: 8.2/8.3 incompatibility
Дата
Msg-id pu63wzenau.fsf@srv.protecting.net
обсуждение исходный текст
Ответ на 8.2/8.3 incompatibility  (Harald Fuchs <hari.fuchs@googlemail.com>)
Список pgsql-general
In article <20080207095031.O22500@megazone.bigpanda.com>,
Stephan Szabo <sszabo@megazone.bigpanda.com> writes:

> On Thu, 7 Feb 2008, Harald Fuchs wrote:
>> This works fine in 8.2.4, but 8.3.0 rejects the ALTER TABLE with the
>> following (somewhat misleading) error message:
>>
>> ERROR:  insert or update on table "t2" violates foreign key constraint "t2_t1id_fk"
>> DETAIL:  Key (t1id)=(t1id1) is not present in table "t1".

> If the types were considered not comparable, you should have gotten a
> message to that effect rather than a not present message.

Yes, this was really confusing.

> More to the point that comparison should have succeeded I think.

Well, it did succeed in 8.2.x, and I'm actually grateful that 8.3.0
noticed the sloppiness on my side.

> What do the following give?

> select * from t1 where id=CAST('t1id1' as VARCHAR(5));

This returns t1id1, as expected.

> select * from ONLY t2 fk LEFT OUTER JOIN ONLY t1 pk ON
>  (pk.id = fk.t1id) WHERE pk.id IS NULL;

This returns an empty result set.

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