Question

我在控制器中有功能，如下所示：

//Create room
public function createRoom(){
    $room = $this->Rooms->newEntity();
    if($this->request->is('post')){
        $room = $this->Rooms->patchEntity($room, $this->request->data);
        $user_field = $this->Rooms->Users->find()->where(['id' => $this->Auth->user('id')]);
        $room->users = [$user_field];
        $room->current = 1;
        $room->owner_id = $this->Auth->user('id');
        Log::debug($room);
        if($this->Rooms->save($room)){
            $this->Flash->success('You have created a new room.');
            return $this->redirect(['action' => 'index']);
        }
        else{
            $this->Flash->error('Error creating new room.');
        }
    }
    $games = $this->Rooms->Games->find('list', ['limit' => 200, 'keyField' => 'id', 'valueField' => 'name']);
    $this->set(compact('room', 'games', 'users'));
    $this->set('_serialize', ['room']);
}

表＆＃34;用户＆＃34;和＆＃34;房间＆＃34;在关联表中连接。当我运行添加一个新的房间，我让用户通过表单选择房间里的人然后保存，但当我想通过$ room-＆gt; users = [$ user_field]自己添加第一个用户时]。它没有被保存。我将$ room对象记录到文件中，两个对象都是相同的（在通过表单添加用户之后以及通过代码添加用户之后）。帮帮我:(

也许我应该在模型中使用beforeSave（）或afterSave（）？

解我找到了一个带有 Code.Working 帮助的解决方案。而不是添加

$this->loadModel('Users');

到initializie（）方法我只是把它放在我的createRoom（）函数中。

最终工作代码如下所示：

//Create room
public function createRoom(){
    $this->loadModel('Users');
    $room = $this->Rooms->newEntity();
    if($this->request->is('post')){
        $room = $this->Rooms->patchEntity($room, $this->request->data);
        $user_field = $this->Users->get($this->Auth->user('id'));
        $room->users = [$user_field];
        $room->current = 1;
        $room->owner_id = $this->Auth->user('id');
        Log::debug($room);
        if($this->Rooms->save($room)){
            $this->Flash->success('You have created a new room.');
            return $this->redirect(['action' => 'index']);
        }
        else{
            $this->Flash->error('Error creating new room.');
        }
    }
    $games = $this->Rooms->Games->find('list', ['limit' => 200, 'keyField' => 'id', 'valueField' => 'name']);
    $this->set(compact('room', 'games', 'users'));
    $this->set('_serialize', ['room']);
}

Answer 1

如果您没有将用户集减少到特定的用户，则find（）的返回值始终是查询对象：Retrieving Data & Results Sets。

尝试使用此方法将检索到的用户集减少到第一个：

$user_field = $this->Rooms->Users->find()->where(['id' => $this->Auth->user('id')])->first();

或者更好，如果＆＃39; id＆＃39;列是您的主键：

// In the initialize() method
$this->loadModel('Users');

// And in the action:
$user_field = $this->Users->get($this->Auth->user('id'));

Answer 2

我想这会对你有所帮助：

FXML StackPane doesn't align children correctly

通过链接，您可以轻松地将用户对象链接到房间对象，这是我认为更美观的解决方案。假设你已正确设置了关联！

祝你好运：）

Answer 3

我认为在控制器中加载模型的另一种方法可能是使用dirty()方法。

在您的情况下，您将在$rooms->dirty('users', true);之后立即插入Log::debug($room);并评论用户的加载模型。这会将“房间”的“用户”关联信号标记为脏，因此会在调用$this->Rooms->save($room)时将其保存。

试一试;）

CakePHP 3 - 保存前的数据关联

3 个答案: