On Thu, Aug 23, 2007 at 10:50:30AM +0200, Raymond O'Connor wrote: > Thanks for the help. > I'm not sure I fully understand your query, but I dont think that would > work because my popularity variable basically ranges from 1 to ~1.5 > million (where a product with popularity 1 is the best selling product > and a popularity with 1.5 million is the worst selling product). Its > similar to an Amazon sales rank if you're familiar with that.
In this case you could use RangeQueries instead on the popularity field, or add a new field that is set according to the popularity, i.e. on a scale from 1 to 10. The idea of my example was to let products with a higher popularity match the higher boosted sub queries, which should lead to a higher ferret score then. Jens -- Jens Krämer webit! Gesellschaft für neue Medien mbH Schnorrstraße 76 | 01069 Dresden Telefon +49 351 46766-0 | Telefax +49 351 46766-66 [EMAIL PROTECTED] | www.webit.de Amtsgericht Dresden | HRB 15422 GF Sven Haubold, Hagen Malessa _______________________________________________ Ferret-talk mailing list [email protected] http://rubyforge.org/mailman/listinfo/ferret-talk
