Thanks for the help teknoid. I've done that, and I think there are
some possibilities there, but they are a little longer than I'd like.
For those who might come after this and not feel like reading the full
situation above, my question in a nut-shell really is:
How do you run a find on a field in a Model that is two associations
away rather than just one?
E.g. - When querying Post which belongsTo Artist which hasMany
Subscription, how do I compare a field in Post to a field in
Subscription? Cake doesn't seem to want to create a second JOIN in
the SQL output.
On Oct 31, 3:28 pm, teknoid <[EMAIL PROTECTED]> wrote:
> That's a lot to read, but I can point you in the direction of checking
> out the Containable behavior (in the manual).
> ... as well as really carefully reading up on the model associations
> and data retrieval.
>
> On Oct 31, 2:30 pm, 33rtp <[EMAIL PROTECTED]> wrote:
>
> > Hey all...
>
> > New to PHP, Cake, and MySQL so bear with me.
>
> > I've been searching high and low for the best way to make this query
> > work, but just haven't gotten it yet.
>
> > Here's my setup:
>
> > I have models for Users, Subscriptions, Authors, and Posts where:
> > User HasMany Subscription (pk_User.id, fk_Subscription.user_id),
> > Author, HasMany Subscription (pk_Author.id,
> > fk_Subscription.author_id),
> > Author HasMany Post (pk_Author.id, fk_Post.author_id)
> > and all of the related BelongsTo's as well.
>
> > I want to search $this->User->Subscription->Author->Post (from the
> > UsersController) for all posts where the logged in user has a current
> > subscription to that(/those) author(s).
>
> > E.g. - ($this->Auth->user('id') = Subscription.user_id AND
> > Subscription.expiration_date >= date ('Y m d') AND
> > Subscription.author_id = Post.author_id).
>
> > Further, each Post contains a second field (INT) called
> > Post.access_level and this must be lower than the value stored in
> > Subscriptions.subscription_level.
>
> > The trick, of course, isn't just doing this, but doing it
> > efficiently. Here are the options I've thought of.
>
> > - I could query $this->User->Subscriptions for relevant subscriptions
> > and return either the full array (set to $subscriptions) or a 'list'
> > of key/value pairs where key => author_id and value =>
> > subscription_level. However, in issuing a second find() call to Post,
> > I don't know how I would compare the key/value pairs in the
> > $subscriptions array with the values for 'Post.author_id' and
> > 'Post.access_level' where the evaluation occurs at the row level in
> > $subscriptions. With the find('all') array I mentioned first,
> > $subscriptions returns an array with [key][Subscription][field] so I
> > can't set conditions for 'Post.author_id' and 'Post.access_level'
> > without a foreach() which I don't want because of the extra database
> > queries it would generate.
>
> > -Alternately, I could use BindModel() to create a HABTM (or I could
> > just create it permanently in the models) relationship between
> > Subscription and Post. This option requires an extra join table in my
> > database though, and may result in slower database queries as all the
> > joins are performed. Additionally, there are other models (File,
> > Event, etc) that are owned by Author and each of these would require
> > an extra join table and HABTM relationship. I could be wrong, but
> > doing all of this seems somehow redundant and there should be a more
> > elegant solution.
>
> > -There are also probably other ways using PHP functions to manipulate
> > the arrays after they have been returned from find calls (e.g.
> > Subscriptions->find w/ conditions and Post->find w/ conditions and
> > then some PHP array functions to compare those two arrays), but I'm
> > too new to this to know where to even start with that.
>
> > There's got to be a simple method I'm missing (or just don't know
> > about yet). Any ideas?
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