Hi folks.
This is my first post here. This Group saved me a lot of times (even
when I wasn't member of it), but these days I've been "googling"
around and didn't found a good solution for my problem.
Where I work, we have some integrated databases. All of them in the
same mysql server, but in different schemas. Our users login with the
same username/password for all systems. (Here we work with two
platforms: php and java).
I've used $useDbConfig to use the configuration for the other
database, that was set in /config/database.php.
The problem is: As I experienced, cakePHP can't handle "joins" between
those databases, because instead use a prefix to identify the correct
database, it just create two different connections, and of course, he
can't join them.
My first solution was create a mysql view in the main database of my
application. In this case, cakePHP was able to retrieve the data in
the same connection and join the tables, but I don't know if it's the
best solution for that problem.
My Question: Is there a solution to join these tables in different
databases just using cakePHP, or will I need to use these mysql
views ?!
OBS: In some cases, we don't use cakephp convention, due to this mix
of integrated databases of another systems that doesn't uses cake
convencion, so I needed to set $useTable, and $primaryKey.
This is my /config/database.php:
var $default = array(
'driver' => 'mysql',
'persistent' => false,
'host' => 'localhost',
'login' => 'root',
'password' => '',
'database' => 'cidadaopg',
'prefix' => '',
);
var $gesed = array(
'driver' => 'mysql',
'persistent' => false,
'host' => 'localhost',
'login' => 'root',
'password' => '',
'database' => 'gesed',
'prefix' => '',
);
This is my Noticia Model (News in portuguese)
class Noticia extends AppModel{
var $name = "Noticia";
var $useTable = "Noticia";
var $primaryKey = "cd_noticia";
var $belongsTo = array(
"CategoriaNoticia" => array(
"className" => "CategoriaNoticia",
"foreignKey" => "cd_categoria",
"type" => "INNER"
),
"Funcionario" => array(
"className" => "Funcionario",
"foreignKey" => "cd_funcionario",
),
);
}
//---------------------------------------------------------------
My Funcionario Model (employee in portuguese)
class Funcionario extends AppModel{
var $name = "Funcionario";
var $useDbConfig = "gesed";
var $useTable = "Funcionario";
var $primaryKey = "cd_funcionario";
}
Thanks for the attention and help.
Thiago Elias.
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