Question

我有3张桌子

证书，用户，状态

他们有这些关系：

certificates - user, one -- many
user-status, one -- one

我也有activeQuery：

$this->andWhere(['<', 'created', new Expression("DATE_SUB(NOW(), INTERVAL " . $days . " DAY)")])
            ->andWhere(['is_active' => 1])
            ->andWhere(['<', 'delayed_until', new Expression("NOW()")])
            ->andWhere(['<', 'resend_count', LeCertificate::RESEND_COUNT])

是否可以添加范围->andWhere，我可以获得status.not_deleted = 0当前证书？

Answer 1

假设您已在LeCerificate和User类中定义了正确的关系：

LeCertificate::find()->alias('C')
    ->joinWith(['users', 'users.status S'], false, 'INNER JOIN')
    ->andWhere(['<', 'C.created', new Expression("DATE_SUB(NOW(), INTERVAL " . $days . " DAY)")])
    ->andWhere(['C.is_active' => 1])
    ->andWhere(['<', 'C.delayed_until', new Expression("NOW()")])
    ->andWhere(['<', 'C.resend_count', LeCertificate::RESEND_COUNT])
    ->andWhere(['S.not_deleted' => 0]);

关系应如下所示：

class LeCertificate extends ActiveRecord 
{
    ...

    public function getUsers()
    {
        return $this->hasMany(User::className(), ['certificate_id' => 'id']);
    }

    ...
}

和

class User extends ActiveRecord 
{
    ...

    public function getStatus()
    {
        return $this->hasOne(Status::className(), ['id' => 'status_id']);
    }

    ...
}

Yii2：ActiveQuery使用关系

1 个答案: