Is *the path* below the same as "search path* in other
postings about this thread ?
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
