Re: Having difficulty writing a "best-fit" query..

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От Harald Fuchs
Тема Re: Having difficulty writing a "best-fit" query..
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Msg-id pulka30xut.fsf@srv.protecting.net
обсуждение исходный текст
Ответ на Having difficulty writing a "best-fit" query..  ("Jamie Tufnell" <diesql@googlemail.com>)
Список pgsql-sql
In article <b0a4f3350710160946y4a31a227sbffbf085cdbfa487@mail.gmail.com>,
"Jamie Tufnell" <diesql@googlemail.com> writes:

> Hi list,
> I have a many-to-many relationship between movies and genres and, in the link
> table I have a third field called which orders the "appropriateness" of the
> relations within each movie. 

> For example:

> movie_id, genre_id, relevance (i've used movie/genre titles for clarity here,
> but in reality they're id's)
> --------------------------------------------
> beverly hills cop, action, 2
> beverly hills cop, comedy, 1
> the heartbreak kid, comedy, 2
> the heartbreak kid, romance, 1

> The above would mean, to my application:
> "Beverly Hills Cop is both an Action and a Comedy movie but primarily an Action
> movie."
> "The Heartbreak Kid is both a Comedy and a Romance movie but primarily a Comedy
> movie."

> First of all, if there's a better way to model this kind of ranking/ordering of
> many-to-many relationships, please let me know.

This looks fine to me.

> Now, to my problem..

> I'm taking a subset of all my genres, and I want to get ONE row for each movie
> in the subset alongside its most appropriate genre (whichever has the highest
> relevance).  In other words, the best fit.

You could use something like that:
 SELECT m.name, g.name, mg.relevance FROM movies m JOIN mg ON mg.movie_id = m.id JOIN genres g ON g.id = mg.genre_id
LEFTJOIN mg mg1 ON mg1.movie_id = mg.movie_id                 AND mg1.relevance > mg.relevance WHERE mg1.movie_id IS
NULL

This means that there must not be an link table entry for the same
movie with a higher relevance.



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