Yii 2：使用保存的数据表的值将数据保存到一个表后填充另一个表

Question

我有两张表：Stations 和 Routes。 station 表有列（station_id、address、office_hours、date_create），Routes 表有列（routes_id、from_id、destination_id）。 Routes 表中的 from_id 和 destination_id 是引用站表的外键。

现在我想做的是每当添加一个车站时，该车站的路线是使用表中已有的车站计算的。例如，假设我们在站表中已经有站 A。将 B 站结果添加到两条路线 A-->B 和 B-->A 其中对于路线 A-->B，A =>from 而 B => 目的地，因此它们的 ID 仅在路线表中被挑选和填充相应地。

我在工作站控制器中尝试了以下代码，但没有成功。代码是：

站控制器：

public function actionCreate()
{
    $model = new Stations();

    $routesModel = new Routes();

    //checking whether we are getting the logged in user id value
    Yii::info("User id=".Yii::$app->user->id); 

    $model->registered_by_id = Yii::$app->user->id;

    $model->status = 10;

    if ($model->load(Yii::$app->request->post()) && $model->save()) {

        //checking here the saved user id value in  table
        Yii::info("checking User id after saving model=".$model->registered_by_id);

        $this->createRoutes($model->station_id);

        return $this->redirect(['view', 'id' => $model->station_id]);
    } else {
        return $this->render('create', [
            'model' => $model,
        ]);
    }
}

public function createRoutes($id)
{
    $model = new Routes;

    $command = Yii::$app->db->createCommand('SELECT station_id FROM stations WHERE station_id !='.$id);
    $all_stations = $command->queryAll();

    foreach ($all_stations as $new_route) {
        $from_id = $id;
        $destination_id = $new_route;
        $model->load(Yii::$app->request->post()) && $model->save();
        $from_id = $new_route;
        $destination_id = $id;
        $model->load(Yii::$app->request->post()) && $model->save();
    }

    return;
}

站表已填充，但路线表未填充。但是，我没有收到任何错误。我哪里出错了？

请尽可能提供帮助。

修改函数createRoutes()

public function createRoutes($id)
{
    $model = new Routes;

    $count = (new \yii\db\Query())->from('stations')->count();

    if($count > 1) {
        $command = Yii::$app->db->createCommand('SELECT station_id FROM stations WHERE station_id !='.$id);
        $all_stations = $command->queryAll();

        foreach ($all_stations as $new_route) {
            $model->from_id = $id;
            $model->destination_id = $new_route['station_id'];
            $model->save();
            $model->from_id = $new_route['station_id'];
            $model->destination_id = $id;
            $model->save();
        }
        return;
    }
    else
        return;
}

现在，在进行上述更改后，它只将一条路线保存到数据库中，但有几个车站，因此应该创建几条路线。如果我有 10 个站，它只会生成站 9 到 10 的路线；我的假设是它只保存 foreach 外观的最后一点，在其他情况下，不会调用 $model->save() 参数。我在这里做错了吗？

Answer 1

在您的 createRoutes 方法中，您只创建了一个 Routes 模型并填充它并将其保存在循环中，这就是您只获得一条记录的原因。相反，您只需要像这样在 foreach 中创建新模型：

public function createRoutes($id)
{
    //$model = new Routes; not here

    // $command = Yii::$app->db->createCommand('SELECT station_id FROM stations WHERE station_id !='.$id);
    // $all_stations = $command->queryAll();

    // I am using  active record instead of single query, you can do anything you want here
    $all_stations = Stations::find()
                ->where('station_id != :id', [':id' => $id])
                ->all();

    foreach ($all_stations as $new_route) {
        $model = new Routes; // create record here
        $model->from_id = $id;
        $model->destination_id = $new_route->station_id;
        // no need to load because these data are not in request param
        // $model->load(Yii::$app->request->post()) && $model->save();
        $model->save();

        $model = new Routes; // and another one
        $model->from_id = $new_route->station_id;
        $model->destination_id = $id;
        $model->save();
    }

    return;
}

---

重构版本： 上面应该可以解决问题，但这里有另一个优化。

在Routes 模型中创建一个像这样的静态方法：

public static function create($data)
{
    $model = new self;
    $model->from_id = $data['from_id'];
    $model->destination_id = $data['destination_id'];
    return $model->save();
}

然后像这样重写createRoutes 方法：

public function createRoutes($id)
{
    $all_stations = Stations::find()
                ->where('station_id != :id', [':id' => $id])
                ->all();

    foreach ($all_stations as $new_route) {
        Routes::create([
            'from_id' => $id,
            'destination_id' => $new_route->station_id,
        ]);

        Routes::create([
            'from_id' => $new_route->station_id,
            'destination_id' => $id,
        ]);
    }

    return;
}

而且我们现在有了更简洁的版本。

Answer 2

也许 Yii::$app->request->post() 在函数 createRoules 中不起作用，因为它不是一个动作函数，但你可以使用 $model->getErrors() 来查找模型中的错误

Answer 3

我知道这很粗略，对系统的查询时间意义重大，但我正在拼命寻找答案。所以到目前为止这对我有用；

public function actionCreate()
{
    $model = new Stations();

    $routesModel = new Routes();

    //checking whether we are getting the logged in user id value
    Yii::info("User id=".Yii::$app->user->id); 

    $model->registered_by_id = Yii::$app->user->id;

    $model->status = 10;

    if ($model->load(Yii::$app->request->post()) && $model->save()) {

        //checking here the saved user id value in  table
        Yii::info("checking User id after saving model=".$model->registered_by_id);

        $this->createRoutes($model->station_id);

        return $this->redirect(['view', 'id' => $model->station_id]);
    } else {
        return $this->render('create', [
            'model' => $model,
        ]);
    }
}

public function createRoutes($id)
{
    $model = new Routes;

    $count = (new \yii\db\Query())->from('stations')->count();

    if($count > 1) {
        $command = Yii::$app->db->createCommand('SELECT station_id FROM stations WHERE station_id !='.$id);
        $all_stations = $command->queryAll();

        foreach ($all_stations as $new_route) {
            $from_id = $id;
            $destination_id = $new_route['station_id'];
            $sql = 'INSERT INTO routes(from_id, destination_id) VALUES('.$from_id.','.$destination_id.')';
            Yii::$app->db->createCommand($sql)->execute();
            // $model->save();
            $from_id = $new_route['station_id'];
            $destination_id = $id;
            $sql = 'INSERT INTO routes(from_id, destination_id) VALUES('.$from_id.','.$destination_id.')';
            Yii::$app->db->createCommand($sql)->execute();
            // $model->save();
        }
        return;
    }
    else
        return;
}

我仍然欢迎更好的解决方案，尤其是使上述解决方案如此高效。谢谢各位。