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Jon Heggland wrote:
> Bob Badour wrote:
>> Jon Heggland wrote: >>> Bob Badour wrote: >>>> Why include them in the referencing relation at all? >>> Pragmatic reasons. For instance as an alternative to an ASSERTION-type >>> join-based constraint, which many(? most? Does anybody know?) DMBSs >>> don't support. >> I don't follow your reasoning. If you simply omitted the dependent >> attributes from the referencing relation, normal foreign key constraints >> would suffice.
Please pardon my confusion about the lingo - isn't a foreign key just a shortcut notation for an ASSERTION-type constraint? Ie., what the system must do is the same for both, eg., ItemGroup{I#} join Item{I#} = ItemGroup{I#} would be the assertion that means the same as ItemGroupFKItem(Item{I#}) and similarly for superkeys.
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. But as long as people like the shortcut notation, I think the example is a good reason for allowing super foreignkeys or foreign superkeys!
p Received on Tue Aug 08 2006 - 12:38:44 CDT
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