Yii 2.0 table alias in SQL query - yii2

Question

I am using Yii2.0 and I have following error when I doing filtering with relationship:
Exception (Database Exception) 'yii\db\Exception' with message
'SQLSTATE[42S22]: Column not found: 1054 Unknown column
'userContact.email' in 'where clause' The SQL being executed was:
SELECT tbl_user.* FROM tbl_user LEFT JOIN tbl_user_contact ON
tbl_user.id = tbl_user_contact.user_id WHERE
userContact.email='me#me.com'
And it is obvious that the table name alias is not given. Following is my code that generate the query above:
Class Files
class User extends ActiveRecord{
public function getUserContacts(){
return $this->hasMany(UserContact::className(), ['user_id' => 'id']);
}
}
class UserContact extends ActiveRecord {
public function getUser(){
return $this->hasOne(User::className(), ['id' => 'user_id']);
}
}
Query
User::find()->joinWith('userContacts', false)
->where(['userContact.email' => $email])
->one();
I follow the instruction given here.
Is there a way to have the alias in the query?

Answer 1

Use method "alias('string')".
User::find()->alias('u')->joinWith(['userContacts' => function($query) use ($email){
$query->alias('uc')->where(['uc.email' => $email])
}])
->one();
Look this API doc

Answer 2

In MySQL, your user table is called tbl_user_contact. However, you are referring to it as userContact, which results in the error.
When adding conditions, you should refer to fields using the actual table name. Here's the proper code:
User::find()->joinWith('userContacts', false)
->where([UserContact::tableName().'.email' => $email])
->one();
You could just replace UserContact::tableName().'.email' with tbl_user_contact.email, but using tableName() is better practice.

Answer 3

ActiveQuery extends of Query, you can use methods of query in ActiveQuery:
$query = \app\models\db\AnuncioConsulta::find();
$query->from(\app\models\db\AnuncioConsulta::tableName() . ' as ac' );
$query->join = [
['INNER JOIN', 'anuncio as a' , ' a.id = ac.anuncio_id AND a.status = 1 '],
['INNER JOIN', 'autor as au' , ' au.id = a.autor_id AND au.agente_inmobiliario = 0 '],
['INNER JOIN', 'usuario as u' , ' u.id = au.object_id '],
];
$query->andWhere('ac.news =:status' , [':status' => 1 ]);
$query->andWhere('ac.delete =:status' , [':status' => 0 ]);
$query->andWhere('u.id =:userId' , [':userId' => Yii::$app->user->id ]);
return new \yii\data\ActiveDataProvider([
'query' => $query,
]);