I'd add "reference_count" property (IntegerProperty(default=0)) to
Photo entity, and increment it each time a reference happens. Then,
call Photo.all().order("-reference_count").fetch(100) to get the top
100 most referenced photos.

A few additional notes:
1. If you need an accurate reference count number, you need to put the
routine (which reads the Photo entity, increment the reference count,
and put it back to the database) into a transaction, but it is
relatively expensive. I usually don't put it in a transaction unless I
absolutely need an accurate number.
2. Making too many put() calls in one HTTP request is not a good idea.
I'd suggest to perform non critical operations (such as incrementing
reference count) asynchronously using Task Queue.

Satoshi

On Jan 6, 4:15 pm, Daniel Aguilar <[email protected]> wrote:
> Hi Satoshi,
>
> thanks for your reply, it really helped.
> In fact I just read this 
> artile:http://code.google.com/appengine/articles/modeling.html
> and have started implementing based on it.
> Still wondering a couple of things, though...
> for instance, in my app I have another class called Collage.
> A collage entity has properties like name, date_created, etc... but also
> layer_0, layer_1 and layer_2, which are references to Photo entities.
> What kind of query could i perform in order to get the most referenced
> Photos?
>
> Thanks again!
>
> need to keep track of how many times
>
>
>
> On Thu, Jan 7, 2010 at 12:51 AM, Satoshi <[email protected]> wrote:
> > First of all, please remember that GAE/database is an Object-database,
> > not a Relational-database. You can design your database with
> > relations, but you will likely hit a roadblock later if you heavily
> > rely on relations (because of the lack of JOIN and the performance
> > problem of nested queries).
>
> > If I were you, I would simply create two models (Artist and Photo),
> > and have the "artist" property on Photo class, which is just a
> > reference to an Artist entity (ReferenceProperty).  If you want to
> > show all the Photos done by a particular Artist, you just need to
> > query it (Photos.all().filter('artist', ...)).
>
> > Alternatively, you could specify the Artist entity as the parent
> > entity of each Photo, which essentially creates an entity group for
> > each Artist - which has pros (transactions) and cons (possible
> > performance hit because of transactions).
>
> > Third alternative is ListProperty, but this is difficult to do it
> > right without putting them in an entity group (which is alternative
> > two)...
>
> > Satoshi
>
> > On Jan 5, 9:52 am, Daniel A <[email protected]> wrote:
> > > Hi there,
>
> > > I just started writting my first app after reading some documentation
> > > and tutorials.
> > > Looks like an exciting platform to develop on, but there're many
> > > things I still have doubts about.
> > > I used to develop in PHP/MySQL, and I quite don't get how should I
> > > proceed in order to get an efficient relational model.
>
> > > To simplify things, I have two kind of entities: Artist and Photo. I
> > > need to define relations one-to-many between instances of these two
> > > entities. That is, an Artist can have many Photos, and a Photo can
> > > only have one Artist.
>
> > > In my table-shaped mind, I would model three kinds of entities
> > > (Artist, Photo, ArtistPhotoRelation). Would this approach be the right
> > > thing in GAE/datastore? Maybe I should avoid the relational entity,
> > > and use multiple-valued properties in the Artist instead?
>
> > > Thanks a lot!
>
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