On Thu, 24 Jan 2002, 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.
Indeed. I find the spec makes the most sense after you 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!
Wow. Thanks for diging this out.
> This doesn't look glaringly wrong to me, so maybe you want to consider it.
> Please note step 2.
It looks fine, and is probably what we should do. Well, I'd do things in a
different order (look only in in-path schemas first for instance), but
that's just trying to optimize the query. :-)
How different are the type coercion rules?
Take care,
Bill
