joinwith - Subconsulta Yii2 en registro activo

yii2 find join

Question

¿Cómo puedo convertir este sql en consulta de registro activa?

SELECT * FROM `base_twitter` WHERE id NOT IN (SELECT base_id from base_followers)

Answer 1

Prueba esta consulta

$models = BaseTwitter::find()->where(''id NOT IN (SELECT base_id from base_followers)'')->all();

Answer 2

Suponiendo que sus modelos se denominen BaseTwitter y BaseFollower consecuencia, esto debería funcionar:

$subQuery = BaseFollower::find()->select(''id'');
$query = BaseTwitter::find()->where([''not in'', ''id'', $subQuery]);
$models = $query->all();

Answer 3

// SELECT * FROM (SELECT * FROM `user` WHERE `active` = 1) `activeusers`;
$subquery = (new /yii/db/Query)->from(''user'')->where([''active'' => true])
$query = (new /yii/db/Query)->from([''activeusers'' => $subquery]);

// subquery can also be a string with plain SQL wrapped in parenthesis
// SELECT * FROM (SELECT * FROM `user` WHERE `active` = 1) `activeusers`;
$subquery = "(SELECT * FROM `user` WHERE `active` = 1)";
$query = (new /yii/db/Query)->from([''activeusers'' => $subquery]);