Re: 2nd Normal Form Question

On 8 feb, 21:08, "David Portas"
<REMOVE_BEFORE_REPLYING_dpor..._at_acm.org> wrote:
> "Jan Hidders" <hidd..._at_gmail.com> wrote in message
>
> news:3c0c26bc-65d3-44b8-bf91-ea32c0b4a917_at_i7g2000prf.googlegroups.com...
>
>
>
> > On 8 feb, 17:39, Bob Badour <bbad..._at_pei.sympatico.ca> wrote:
> >> gamehack wrote:
> >> > Hi all,
>
> >> > I'm currently evaluating whether a relation is in 2NF. The relation is
> >> > defined as follows:
> >> > <Year | Winner Name | Winner Votes | Party | Home State> in the
> >> > context of an election. I've given a sample relation below.
> >> > 1946 | MyName | 453 | MyParty | California
> >> > The primary key for this relation is 'Year'.
>
> >> > Now the question is whether this relation is in 2NF? What confuses me
> >> > is that some books say the following:
> >> > "Note that when a 1NF table has no composite candidate keys (candidate
> >> > keys consisting of more than one attribute), the table is
> >> > automatically in 2NF."
>
> >> I am not sure where you read that. It sounds like a typo or a mistake.
> >> Composite keys are important at the higher normal forms.
>
> > Of course, but all that it says is that if you have determined all the
> > candidate keys and they happen all to be not-composite then a 1NF is
> > always also in 2NF. I'm sure you agree that this is correct.
>
> Here's a counter example. PatientId is the candidate key but the relation is
> not in 2NF. This is an unusual case but I don't think it should be ignored.

Yeah, yeah, yeah. I know, I know. I wanted to give Brian Selzer the pleasure of correcting me on this twice, but you beat him to it. :-)

• Jan Hidders