I am facing same problem

On Sep 21, 5:56 pm, Rick <[EMAIL PROTECTED]> wrote:
> I tried that technique but it did not work.  It appears that you can't
> havemultiple"useDBConfig"s in a single query.  This seems reasonable
> since you also can't do a raw select usingmultipleconnections which
> is what the useDBConfigs give you (I think).
>
> Richard
>
> On Sep 20, 5:56 pm, Adam Royle <[EMAIL PROTECTED]> wrote:
>
> > What you need to do is set $useDBConfig for each model.
>
> > var $useDBConfig = 'A';
>
> > then in your DATABASE_CONFIG you would use
>
> >         var $A = array(
> >                 'driver' => 'mysql',
> >                 'persistent' => false,
> >                 'host' => 'localhost',
> >                 'login' => 'username',
> >                 'password' => 'password',
> >                 'database' => 'A',
> >                 'prefix' => '',
> >         );
>
> > Adam
>
> > On Sep 21, 4:24 am, Rick <[EMAIL PROTECTED]> wrote:
>
> > > I've got an existing MySql instance that contains 4 databases (lets
> > > call them A, B, C & D).  Each has tables as you would expect.  For
> > > this post lets say tables are named like table1, table2 etc.. in each
> > >database.  My single user/password has access to all databases in the
> > > instance.
>
> > > My problem is I need to query across several databases.  This is the
> > > way I would do this with a raw query:
>
> > > select * from A.table1 as t1, B.table2 as t2
> > > where t1.id = t2.id
>
> > > I haven't been able to do this through any combination of model
> > > attributes.  Naturally I thought that
>
> > > var $useTable = "A.table1"
>
> > > in The table1 model and
>
> > > var $useTable ="B.table2"
>
> > > in the table2 model would work - alas no it does not.
>
> > > I've also tried using the prefix attribute and that doesn't do it
> > > either.
>
> > > Is there some way to do this in Cake or do I need to forget the models
> > > and just do raw sql queries in the controllers?  It seems to me this
> > > is just an arbitrary restriction in the model code.
>
> > > Richard

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