Re: Jan's well-defined view updates definition

"Jan Hidders" <jan.hidders_at_ua.ac.be> wrote in message news:3f696e04.0_at_news.ruca.ua.ac.be...
> Bob Badour wrote:
> > "Jan Hidders" <jan.hidders_at_pandora.be> wrote in message
> > news:d14ab.25480$rw3.1352807_at_phobos.telenet-ops.be...
> >>Mikito Harakiri wrote:
> >>>"Jan Hidders" <jan.hidders_at_pandora.be> wrote:
> >>>
> >>>> Finally, we define a view V as *updatable* by a certain set of
> >>>> updates U if for all databases D that satisfy the schema all
> >>>> updates in U are well-defined. Additionally we say that V is
> >>>> *commutatively updatable* by U if for all databases D that
> >>>> satisfy the schema it holds that if two series of updates from
> >>>> U have the same result when applied to Q(V) and perform only
> >>>> well-defined updates then they result in the same database when
> >>>> applied to D.
> >>>
> >>> Could you please provide a non-commutatively updatable example
> >>> that we can discuss?
> >>
> >> Lets say I have base relations R(a,b) and S(b,c) with a foreign key
> >> R.b -> S.b and a view V that is defined by the natural join of R
> >> and S. The additions and deletions are both well-defined but if I
> >> add the tuple (a1,b1,c1) and then remove it then the end result is
> >> an additional tuple (b1,c1) in S, which is not the same as the end
> >> result of adding 0 tuples. So for the class of updates that
> >> consists of inserts and deletes it is not commutatively updatable.
> >
> > Why would one have (b1,c1) in S instead of (a1,b1) in R or neither?

>

Since the user instructed the dbms to delete the tuple with c1, I do not see how it violates the CWA. I do, however, see how it ignores symmetry. Received on Thu Sep 18 2003 - 14:51:17 CEST