Re: Having difficulty writing a "best-fit" query..
От | Harald Fuchs |
---|---|
Тема | Re: Having difficulty writing a "best-fit" query.. |
Дата | |
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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