Illuminate\Database\Eloquent\Model::fromJson PHP Метод

fromJson() публичный Метод

Decode the given JSON back into an array or object.
public fromJson ( string $value, boolean $asObject = false ) : mixed
$value string
$asObject boolean
Результат mixed
    public function fromJson($value, $asObject = false)
    {
        return json_decode($value, !$asObject);
    }
Model