Model with variable database

Yii 2.0
1

I need my model to use different databases. I want to pass $company as an argument when instatiating.

class FatFaturas extends \yii\db\ActiveRecord
{

// override of getDB method..
public static function getDb()
{
    return \common\skeyra\utils::get_client_connection($company);
}

I would like to set $company to a value when creating a new object of FatFaturas, but I can´t use $this within a static function. And the function must be static because i’m overriding a method of Activerecord.

How can I accomplish this? Any help would be appreciated.

2

Database connection is required before instance of model is available (you need to have connection to fetch records from database). How do you want to pass $company for FatFaturas::findOne($id)?

3

Thank you rob006.
That’s my problem.
How do I pass $company in order to make it work?
I’ve tried setting a public property. I’ve tried to play with __construct method… nothing worked so far…

4

For those who may have the same problem:

I got it done like this:

class FatFaturas extends \yii\db\ActiveRecord
{
public static $company;

public function __construct($company)
{
        self::$company = $company;
}

// override of activerecord  getDB method
public static function getDb()
{
    return \common\skeyra\utils::get_client_connection(self::$company);
}

get_client_connection is just some code that returns a connection

model shall be instantiated like this:

$something= new FatFaturas(5); e.g. for company number 5

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