Yii2 Delete Query Using Relations - php

Question

I have defined a relation in Yii2 in which every website can have multiple feeds:
public function getFeeds() {
return $this->hasMany(Feed::className(), ['website_id' => 'id']);
}
Now consider following query:
$ids = [2, 3];
$feeds = Website::find()->where(['user_id' => 1])->with(['feeds' => function($query) use ($ids) {
$query->where(['id' => $ids);
}])->all();
The corresponding raw SQL queries are:
SELECT * FROM `website` WHERE `user_id`= 1 // returns 4, 5, 6
SELECT * FROM `feed` WHERE (`id` IN (2, 3)) AND (`website_id` IN (4, 5, 6))
My questions is, can I have delete version of this query using Active Record? Something like this:
SELECT * FROM `website` WHERE `user_id`= 1
DELETE FROM `feed` WHERE (`id` IN (2, 3)) AND (`website_id` IN (4, 5, 6))
Any help would be appreciated.

Answer 1

You may do as this:
$query = Website::find()
->select('id')
->where(['user_id' => 1]);
Feed::deleteAll([
'id' => [2, 3],
'website_id' => $query
]);

Answer 2

In this case could be better a CreateCommand
for do this you should bild raw query using join eg:
select *
from `feed`
join `website` on `website`.id = `feed`.`website_id`
WHERE website`.`user_id`= 1
AND `feed`.id IN (2, 3)
or for delete
delete
from `feed`
join `website` on `website`.id = `feed`.`website_id`
WHERE website`.`user_id`= 1
AND `feed`.id IN (2, 3)
then for Yii2
$sql = "delete
from `feed`
join `website` on `website`.id = `feed`.`website_id`
WHERE website`.`user_id`= 1
AND `feed`.id IN (2, 3);"
Yii::$app->db->createCommand($sql)->execute();