Odd plpgsql behaviour

[Christopher Kings-Lynne]

On 7.4:

This is what we wanted to do:

IF TG_OP = 'INSERT' OR (TG_OP = 'UPDATE' AND NEW.name != OLD.name) THEN
EXECUTE x;
END IF;

However, we had to write it like this:

IF TG_OP = 'INSERT' THEN
EXECUTE x;
ELSIF TG_OP = 'UPDATE' AND NEW.name != OLD.name THEN
EXECUTE x;
END IF;

Because in the first case it would complain that OLD.name wasn't
defined, if the trigger was NOT an update.

OK, but the second case works??!?! Is this a weird peculiarity of the
pl/pgsql lazy evaluation rules? Why doesn't the first one work if the
second one does?

Chris

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[Mike Rylander]

On Mon, 15 Nov 2004 15:12:24 +0800, Christopher Kings-Lynne
<chr...@familyhealth.com.au> wrote:
> On 7.4:
>
> This is what we wanted to do:
>
> IF TG_OP = 'INSERT' OR (TG_OP = 'UPDATE' AND NEW.name != OLD.name) THEN
> EXECUTE x;
> END IF;
>
> However, we had to write it like this:
>
> IF TG_OP = 'INSERT' THEN
> EXECUTE x;
> ELSIF TG_OP = 'UPDATE' AND NEW.name != OLD.name THEN
> EXECUTE x;
> END IF;
>
> Because in the first case it would complain that OLD.name wasn't
> defined, if the trigger was NOT an update.
>
> OK, but the second case works??!?! Is this a weird peculiarity of the
> pl/pgsql lazy evaluation rules? Why doesn't the first one work if the
> second one does?

IIRC, the reason for this is that the entire IF test is passed to the
SQL engine as a SELECT statement after replacing the TG_* identifiers
with their respective values.

Your first example is essentially

IF (SELECT (TG_OP = 'INSERT' OR (TG_OP = 'UPDATE' AND NEW.name !=
OLD.name) IS TRUE) ...

In this case, since OLD.name does not exist during INSERT it cannot be
replaced. Perhaps someone else can shed a little more light on this.

--
Mike Rylander
mryl...@gmail.com
GPLS -- PINES Development
Database Developer

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[Tom Lane]

Mike Rylander <mryl...@gmail.com> writes:
> IIRC, the reason for this is that the entire IF test is passed to the
> SQL engine as a SELECT statement after replacing the TG_* identifiers
> with their respective values.

> In this case, since OLD.name does not exist during INSERT it cannot be
> replaced. Perhaps someone else can shed a little more light on this.

Nope, that's about all there is to say about it ...

regards, tom lane

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