Polymorphically find model from database in Yii2

Question

I have one table in the database(mysql). But this table stores several slightly different types of rows. The type depends on this tables's type column. I have an abstract ActiveRecord class for a table and several descendant subclasses implementing slightly different logic for the rows of the same table of different types. Now I am implementing an update controller action for all the types of rows. I am provided with an id of the row and need to create an ActiveRecord instance representing the row with this id. But I somehow need to create instances of different subclasses depending on the type of the corresponding row.

If I were provided with both a type and an id I could've used a factory to pick a corresponding subclass. But I can already have the type in the database and an id gives me enough information to pick it from there. But if I were to pick the type from the database first and then to create an instance of the corresponding subclass that would've meant executing the same query twice.

I want to find a good way to get the data from the database and then pick a right ActiveRecord subclass to create an instance for it without making excessive queries or requiring excessive data. Is there a way to do it Yii2?

Or should I approach this problem somehow differently? The actual problem is having several almost the same but a bit different entities stored in a single table with a bit different business-logic.

arogachev · Accepted Answer · 2015-03-20 15:58:58Z

7

One of approaches to this problem is called "Single table inheritance" and described by Martin Fowler here. There is also good article about its implementation in Yii 2 in samdark's (one of the main Yii 2 contributors) cookbook, which is currently in process of writing but is available on Github.

I'm not going to copy the whole article, but leaving just link is also not enough. Here are some important things:

1) Create common query for all types of objects (for example cars):

namespace app\models;

use yii\db\ActiveQuery;

class CarQuery extends ActiveQuery {
    public $type;

    public function prepare($builder)
    {
        if ($this->type !== null) {
            $this->andWhere(['type' => $this->type]);
        }
        return parent::prepare($builder);
    }
}

2) Create separate model for each type (extending from common model Car):

Sport cars:

namespace app\models;

class SportCar extends Car
{
    const TYPE = 'sport';

    public static function find()
    {
        return new CarQuery(get_called_class(), ['type' => self::TYPE]);
    }

    public function beforeSave($insert)
    {
        $this->type = self::TYPE;
        return parent::beforeSave($insert);
    }
}

Heavy cars:

namespace app\models;

class HeavyCar extends Car
{
    const TYPE = 'heavy';

    public static function find()
    {
        return new CarQuery(get_called_class(), ['type' => self::TYPE]);
    }

    public function beforeSave($insert)
    {
        $this->type = self::TYPE;
        return parent::beforeSave($insert);
    }
}

3) Override instantiate() method in Car model to return correct type of cars:

public static function instantiate($row)
{
    switch ($row['type']) {
        case SportCar::TYPE:
            return new SportCar();
        case HeavyCar::TYPE:
            return new HeavyCar();
        default:
           return new self;
    }
}

Then you can use any type of cars individually as regular models.

answered Mar 20, 2015 at 15:58

arogachev

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2 Comments

Gherman Over a year ago

So overriding instantiate lets me choose from the different ActiveRecord subclasses according to the data retrieved from the database. That is exactly what I was asking about! That way I won't need to provide the type to my controller when working with the already existing row. Great thanks!

Olle Härstedt Over a year ago

instantiate is hard to type-check in Psalm. Might be better to use a separate factory class instead.

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