Re: Jan's well-defined view updates definition
Date: Thu, 18 Sep 2003 21:35:39 -0400
Message-ID: <_Duab.217$GP1.20310112_at_mantis.golden.net>
"Jan Hidders" <jan.hidders_at_pandora.be> wrote in message
news:aNoab.27111$OG5.1550428_at_phobos.telenet-ops.be...
> Bob Badour wrote:
> > "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...
> >> >>
> >> >> 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?
> >>
> >> The first option violates the foreign key and the second option deletes
> >> more than necessary which violates the CWA.
> >
> > Since the user instructed the dbms to delete the tuple with c1, I do not
> > see how it violates the CWA.
>
> The tuple (a1,b1,c1) in V is not equal to the tuple (b1,c1) in S. They are
> different values and represent different (although related) facts. Since
> the DBMS was told in the past that the fact represented by (b1, c1) in S
> was true and now is neither directly nor indirectly (it does not logically
> follow) told that it is false, it should assume that it is still true.
>
> > I do, however, see how it ignores symmetry.
>
> There is no symmetry in this case, the foreign key breaks it.
The foreign key is nothing but a short-hand for a WFF. How does a WFF alter symmetry? Received on Fri Sep 19 2003 - 03:35:39 CEST