Re: Jan's well-defined view updates definition
Date: Thu, 18 Sep 2003 20:42:46 GMT
Message-ID: <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.
> I do, however, see how it ignores symmetry.
There is no symmetry in this case, the foreign key breaks it.
- Jan Hidders