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

我有两个表格：工作站和路由。该工作站表具有列（station_id，address，office_hours，date_create），Routes表包含列（routes_id，from_id，destination_id）。路由表上的from_id和destination_id是引用站点表的外键。

现在我想要做的是无论何时添加一个电台，电台的路线都是使用已经在电台中的电台计算出来的。例如，假设我们已经在车站表中有车站A.将站点B的结果添加到两个路由中，A→B和B→A，从而路由A→B，A =>而B =>目的地，因此他们的ID仅在路由表中被挑选和填充因此。

我已经在车站控制器中尝试了以下代码，但是我没有收到任何成功。该代码是：

StationsController：

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; 
} 

该站表被填充，但是没有填充路由表。但是，我没有收到任何错误。我哪里可能会出错？

请尽可能地协助您。

修订功能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，让上述更改后，只保存到数据库的单一路线，但有几个站，因此几条路应创建。如果我有10个电台，它只会为9到10电台生成路由;我的假设是它只保存了foreach外观的最后一部分，在其他情况下，不会调用$ model-> save（）参数。我在这里做错了什么？

Answer 1

在您的createRoutes方法中，您只创建一个Routes模型并填充并将其保存在循环中，这就是为什么您只能获取一条记录的原因。相反，你只需要像这样创建内部的foreach新模式：

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： 以上应该做的伎俩，但这里是另一种优化。

在 Routes模型

创建一个静态方法是这样的：

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

然后重写createRoutes方法是这样的：

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; 
} 

而且我们有很多更清洁的版本了。

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，但是您可以使用$ model-> getErrors（）来查找错误模型

Answer 3

我知道这是如此粗糙，意味着很多系统在查询时间方面，但我拼命寻找答案。所以到目前为止，这对我来说已经奏效。

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，特别是使上述如此高效。感谢帅哥。