php - Yii2 在 Controller 中使用模型中的方法(REST API)

Question

我用 Yii2 框架制作了一个简单的 REST API。我制作了一个带有一些查询的 Controller 以从数据库中获取数据。但现在我想将这些查询与 Controller 分开。我认为我应该将获取查询的函数放在模型 (ActiveRecord) 中，但我无法从我的 Controller 访问该方法。

namespace app\models;

use yii\db\ActiveRecord;

class Car extends ActiveRecord {
    public function __construct(){
        parent::__construct();
    }

    public static function tableName(){
        return 'auto_new';
    }

    public static function primaryKey(){
        return ['auto_id'];
    }

    // Regels voor POST callback
    public function rules(){
        //voorbeeld
        /*
        return [
            [['username', 'password'], 'required']
        ];
        */
    }

    public static function getCar(){
        $query = (new Query())
            ->select([
                'a.auto_id                          auto_id',
                'm.merk                             merk',
                'a.model                            model',
                'a.uitvoering                       uitvoering',
                'k.kleur                            kleur',
                't.transmissie                      transmissie',
                'a.kmstand                          kmstand',
                'a.bouwjaar                         bouwjaar',
                'a.vermogen                         vermogen',
                'b.brandstof                        brandstof',
                'g.garantie                         garantie',
                'a.prijs                            prijs',
                'kl.postcode                        postcode',
                'kl.woonplaats                      woonplaats',
                'GROUP_CONCAT(DISTINCT(atn.NL))     opties',
                'GROUP_CONCAT(DISTINCT(f.foto_id))  fotos'
            ])

            ->from('auto_new a')

            ->join('INNER JOIN', 'tbl_merken m', 'a.merk = m.merk_id')                          //Merk
            ->join('INNER JOIN', 'tbl_kleur k', 'a.kleur = k.kleur_id')                         //Kleur
            ->join('INNER JOIN', 'tbl_transmissie t', 'a.transmissie = t.transmissie_id')       //Transmissie
            ->join('INNER JOIN', 'tbl_brandstof b', 'a.brandstof = b.brandstof_id')             //Brandstof
            ->join('INNER JOIN', 'tbl_garantie g', 'a.garantie = g.garantie_id')                //Garantie
            ->join('INNER JOIN', 'klanten kl', 'a.ac = kl.ac')                                  //Klantgegevens

            //Alle opties van de auto
            ->leftJoin('auto_accessoire acc', 'a.auto_id = acc.auto_id AND a.ac = acc.ac')
            ->leftJoin('tbl_accessoires_trader_new atn', 'acc.code_id = atn.code_id')
            //Alle foto's van de auto
            ->leftJoin('auto_foto f', 'a.auto_id = f.auto_id AND a.ac = f.ac')

            ->groupBy('a.auto_id')
            ->all();
        return $query;
    }
}

这是我 Controller 中的方法:

public function actionGetcarsbyac($ac) {
    Yii::$app->response->format = Response::FORMAT_JSON;
    $model = $this->modelClass;
    $query = Car::getCar();
    $query->where(['a.ac' => $ac]);
    return $query;
}

当我尝试打开 url 时，出现以下错误:

Call to a member function getUniqueId() on a non-object

我不知道要做什么或将这些查询方法放在哪里。

Answer 1

我认为这是因为当您输入 ->all() 或 ->one() 或任何其他 query method它为您提供了无法进一步查询的填充结果对象...

因此，当您想进一步处理该查询时，您应该返回不带查询方法的查询，然后在您的 Controller 操作中完成该查询。所以通过这样做你的代码将是这样的:

public static function getCar(){
        $query = new Query();
            $query->select([
                'a.auto_id                          auto_id',
                'm.merk                             merk',
                'a.model                            model',
                'a.uitvoering                       uitvoering',
                'k.kleur                            kleur',
                't.transmissie                      transmissie',
                'a.kmstand                          kmstand',
                'a.bouwjaar                         bouwjaar',
                'a.vermogen                         vermogen',
                'b.brandstof                        brandstof',
                'g.garantie                         garantie',
                'a.prijs                            prijs',
                'kl.postcode                        postcode',
                'kl.woonplaats                      woonplaats',
                'GROUP_CONCAT(DISTINCT(atn.NL))     opties',
                'GROUP_CONCAT(DISTINCT(f.foto_id))  fotos'
            ])

            ->from('auto_new a')

            ->join('INNER JOIN', 'tbl_merken m', 'a.merk = m.merk_id')                          //Merk
            ->join('INNER JOIN', 'tbl_kleur k', 'a.kleur = k.kleur_id')                         //Kleur
            ->join('INNER JOIN', 'tbl_transmissie t', 'a.transmissie = t.transmissie_id')       //Transmissie
            ->join('INNER JOIN', 'tbl_brandstof b', 'a.brandstof = b.brandstof_id')             //Brandstof
            ->join('INNER JOIN', 'tbl_garantie g', 'a.garantie = g.garantie_id')                //Garantie
            ->join('INNER JOIN', 'klanten kl', 'a.ac = kl.ac')                                  //Klantgegevens

            //Alle opties van de auto
            ->leftJoin('auto_accessoire acc', 'a.auto_id = acc.auto_id AND a.ac = acc.ac')
            ->leftJoin('tbl_accessoires_trader_new atn', 'acc.code_id = atn.code_id')
            //Alle foto's van de auto
            ->leftJoin('auto_foto f', 'a.auto_id = f.auto_id AND a.ac = f.ac')

            ->groupBy('a.auto_id')
            return $query;
    }

请注意，我已从 getCar() 中删除了 ->all()。现在你的 Action 会是这样的:

public function actionGetcarsbyac($ac) {
    Yii::$app->response->format = Response::FORMAT_JSON;
    $model = $this->modelClass;
    $query = Car::getCar();
    $query->where(['a.ac' => $ac])
    ->all(); // here you are completing your query
    return $query;
}

希望这能奏效...