On Mon, 28 Jan 2002, Hiroshi Inoue wrote:
> Is *the path* below the same as "search path* in other
> postings about this thread ?
I think so. I believe the path I've been talking about is the one in step
6 below.
Take care,
Bill
> Maybe Peter's posting isn't the one exactly what I have to
> ask but there are too many postings for me to follow.
>
> regards,
> Hiroshi Inoue
>
> Peter Eisentraut wrote:
> >
> > Bill Studenmund writes:
> >
> > > Does SQL'99 say anything about this?
> >
> > Yes, though, as usual, you have to twist your brain a little to understand
> > it. My understanding is that for a function call of the form "foo(a, b)"
> > it goes like this:
> >
> > 1. Find all functions named "foo" in the current database. This is the
> > set of "possibly candidate routines".
> >
> > 2. Drop all routines that you do not have EXECUTE privilege for. This is
> > the set of "executable routines".
> >
> > 3. Drop all routines that do not have compatible parameter lists. This is
> > the set of "invocable routines".
> >
> > 4. Drop all routines whose schema is not in the path. This is the set of
> > "candidate routines".
> >
> > 5. If you have more than one routine left, eliminate some routines
> > according to type precedence rules. (We do some form of this, SQL99
> > specifies something different.) This yields the set of "candidate subject
> > routines".
> >
> > 6. Choose the routine whose schema is earliest in the path as the "subject
> > routine".
> >
> > Execute the subject routine. Phew!
> >
> > This doesn't look glaringly wrong to me, so maybe you want to consider it.
> > Please note step 2.
> >
> > --
> > Peter Eisentraut peter_e@gmx.net
>
