Re: Foreign superkey support
From: paul c <toledobythesea_at_oohay.ac>
Date: Tue, 08 Aug 2006 18:28:55 GMT
Message-ID: <HJ4Cg.349208$iF6.31975_at_pd7tw2no>
>
> I don't think of constraints as being part of the algebra per se.
> ...
Date: Tue, 08 Aug 2006 18:28:55 GMT
Message-ID: <HJ4Cg.349208$iF6.31975_at_pd7tw2no>
Marshall wrote:
> paul c wrote:
>> Perhaps all I'm saying is that the basic relational algebra doesn't need >> a foreign key concept which I admit is a deviation from the specific >> question.
>
> I don't think of constraints as being part of the algebra per se.
> ...
If they aren't, I think that would mean the algebra is incomplete (and every relation's value could only get 'larger', until it reached the product of its domains' values which could then never be 'constrained' to a smaller set).
p Received on Tue Aug 08 2006 - 20:28:55 CEST