Yii2 How can i set the db config in new Query without use all(), one() or other execution method?

Question

I have 2 models on my search function joined by a field an I want to create an ActiveDataProvider for my GridView. This is what I have so far:

My yii db config:

db => users logs data;
db1 => system data;

Model1 and Model2 are in db1

public function search($params)
{
    // (...)
    $query1 = Model1::find();
    $query2 = Model2::find();
    $query1->select(
            new Expression ("'Data model1: ' Model"),
            'model1.id',
            'model1.columns',
       )->jointWith(['mODEL2'])
       ->andWhere(new Expression ('model1.id || model1.columns != model2.id || model2.columns'));

    $query2->select(
            new Expression ("'Data model2: ' Model"),
            'model2.id',
            'model2.columns',
        )->jointWith(['mODEL1'])
        ->andWhere(new Expression ('model1.id || model1.columns != model2.id || model2.columns'));

    $unionQuery = (new  Query())->from(['dummy_name'=>$query1->union($query2)]);

    $dataProvider = new ActiveDataProvider([
         'query'=>$unionQuery,
    ]);
    // (...)

    return $dataProvider;
}

And the server database exception:

the table or view not exist

In my models (1 and 2) I do specific that I use the db1 connection, but how can I do that in the $unionQuery without use $unionQuery->all(Yii::$app->db1)?

That do not work.

Answer 1

In your case query is executed at ActiveDataProvider level so you should set ActiveDataProvider::$db property with correct db connection.

awesome thanks a lot this works

$dataProvider = new ActiveDataProvider([
    'query'=>$unionQuery,
    'db'=>Yii::$app->db1,      
]);