YII2 creating relations in models between tables from 2 databases - yii2

Question

I have defined 2 databases , for example
return [
'components' => [
'db1' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=db1name',
'username' => 'db1username',
'password' => 'db1password',
],
'db2' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=db2name',
'username' => 'db2username',
'password' => 'db2password',
],
],
];
Now i have a table as 'users' in 'db1' and table 'countries' in 'db2'
users
id , country_code , username , password
1 , DE , xyz , 12345
2 , FR , abc , 12345
countries
code , name
DE , Germany
FR , France
IN , India
I have defined the foreign key relation between users.country_code & countries.code
ISSUE
But when i try to create the model for 'users' table using gii it gives an error , possibly because the tables relation are from 2 different databases.
How to use tables from different databases in relations of a model.
Any suggestions are welcomed

Answer 1

This works in my case to list iten on GridView::widget
-> bd_sisarc is my secound data base
-> deposito_sondagem is a table from my first data base
public static function getDb() // on your model
{
return Yii::$app->get('db1');
}
public static function getDb() // on your model
{
return Yii::$app->get('db2');
}
public function getEmpresaSondagem() // Relation on you model
{
return $this->hasOne(EmpresaSondagem::className(), ['idEmpSondagem' => 'entidade_deposito']);
}
public function search($params)
{
$this->load($params);
$sql = "SELECT deposito_sondagem.*
FROM
deposito_sondagem,
`bd_sisarc`.`tbempresasondagem`
WHERE
`bd_sisarc`.`tbempresasondagem`.`idEmpSondagem`=`deposito_sondagem`.`entidade_deposito`
and deposito_sondagem.estado=1
and tbempresasondagem.estado=1
and numero_registo LIKE '%$this->numero_registo%'
and nomeempsondagem LIKE '%$this->nomeEntidade%'
and dono_sondagem LIKE '%$this->dono_sondagem%'
and data_deposito LIKE '%$this->data_deposito%'";
$query = DepositoSondagem::findBySql($sql);
$dataProvider = new ActiveDataProvider([
'query' => $query,
]);
if (!$this->validate()) {
// uncomment the following line if you do not want to return any records when validation fails
// $query->where('0=1');
return $dataProvider;
}
return $dataProvider;
}

Answer 2

Try this one
SELECT `users`.* FROM `users` LEFT JOIN `db2name`.`countries` ON `users`.`country_code` = `db2name`.`countries`.`code `