Yii如何以一种形式修改两个数据库表

Question

我已经创建了一个表单来从我的网站页面编辑我的数据库中的一个表（hotel_rooms），它可以正常工作。 但是我需要在 privious 表 (hotel_rooms) 的编辑网页中包含另一个表 (hotel_rooms_checked) 中的一些字段，而不是所有表 请问我该怎么做？其中hotel_rooms.Id = hotel_rooms_checked.Id 这是表单的代码

    <?php
    /* @var $this HotelRoomsController */
    /* @var $model HotelRooms */
    /* @var $form CActiveForm */
    ?>

    <div class="form">

    <?php $form=$this->beginWidget('CActiveForm', array(
        'id'=>'hotel-rooms-form',
        'enableAjaxValidation'=>false,
    )); ?>

        <p class="note">Fields with <span class="required">*</span> are required.</p>

        <?php echo $form->errorSummary($model); ?>

        <div class="row">
            <?php echo $form->labelEx($model,'Id'); ?>
            <?php echo $form->textField($model,'Id',array('size'=>60,'maxlength'=>255)); ?>
            <?php echo $form->error($model,'Id'); ?>
        </div>

.......
    <?php $this->endWidget(); ?>

    </div><!-- form -->

这是控制器代码

public function actionUpdate($id)
    {
        $model=$this->loadModel($id);

        // Uncomment the following line if AJAX validation is needed
        // $this->performAjaxValidation($model);

        if(isset($_POST['HotelRooms']))
        {
            $model->attributes=$_POST['HotelRooms'];
            if($model->save())
        // $this->redirect(Yii::app()->request->urlReferrer);

                $this->redirect(array('hotels/index','id'=>$model->Id));
        }

        $this->render('update',array(
            'model'=>$model,
        ));
    }

查看代码是

<?php
/* @var $this HotelRoomsController */
/* @var $model HotelRooms */

$this->breadcrumbs=array(
    'Hotel Rooms'=>array('index'),
    $model->Id,
);

$this->menu=array(
    array('label'=>'List HotelRooms', 'url'=>array('index')),
    array('label'=>'Create HotelRooms', 'url'=>array('create')),
    array('label'=>'Update HotelRooms', 'url'=>array('update', 'id'=>$model->Id)),
    array('label'=>'Delete HotelRooms', 'url'=>'#', 'linkOptions'=>array('submit'=>array('delete','id'=>$model->Id),'confirm'=>'Are you sure you want to delete this item?')),
    array('label'=>'Manage HotelRooms', 'url'=>array('admin')),
);
?>

<h1>View Room : <?php echo $model->Id; ?></h1>

<?php $this->widget('zii.widgets.CDetailView', array(
    'data'=>$model,
    'data'=>$model2,
    'attributes'=>array(
        'Id',
        'HotelName',
        'HotelRoomsType',
        ....
    ),
)); ?>

桌子的结构

Hotel_Rooms
Id
HotelName
HotelRoomsType
.....

Hotel_Rooms_Checked
Id
DateCheckedIn
DateCheckedOut
....

Answer 1

嗯，这个问题似乎还不够清楚。不过，我会尝试猜测您想要什么。

对于view，只需添加其他model 的字段（我想您已经为hotel_rooms_checked 创建了一个模型）

<?php
    /* @var $this PropUnitController */
    /* @var $model HotelRooms */
    /* @var $model2 HotelRoomsChecked */
    /* @var $form CActiveForm */
    ?>

    <div class="form">

    <?php $form=$this->beginWidget('CActiveForm', array(
        'id'=>'hotel-rooms-form',
        'enableAjaxValidation'=>false,
    )); ?>

        <p class="note">Fields with <span class="required">*</span> are required.</p>

        <?php echo $form->errorSummary(array($model, $model2)); ?>

        <div class="row">
            <?php echo $form->labelEx($model,'Id'); ?>
            <?php echo $form->textField($model,'Id',array('size'=>60,'maxlength'=>255)); ?>
            <?php echo $form->error($model,'Id'); ?>
        </div>

        <!--for example, let's say we have an attribute `status` for HotelRoomsChecked-->
        <div class="row">
            <?php echo $form->labelEx($model2,'status'); ?>
            <?php echo $form->textField($model2,'status',array('size'=>60,'maxlength'=>255)); ?>
            <?php echo $form->error($model2,'status'); ?>
        </div>

.......
    <?php $this->endWidget(); ?>

    </div><!-- form -->

现在对于控制器，您可以在实际保存之前对两个模型进行验证。

public function actionUpdate($id)
    {
        $model=$this->loadModel($id);
        $model2 = HotelRoomsChecked::model()->findByPk($id);

        // Uncomment the following line if AJAX validation is needed
        // $this->performAjaxValidation($model);

        if(isset($_POST['HotelRooms']) && isset($_POST['HotelRoomsChecked']))
        {
            $model->attributes=$_POST['HotelRooms'];
            $model2->attributes = $_POST['HotelRoomsChecked'];
            $valid = $model->validate();
            $valid = $valid && $model2->validate();

            if($valid) {
        // $this->redirect(Yii::app()->request->urlReferrer);
                $model->save();
                $model2->save();
                $this->redirect(array('hotels/index','id'=>$model->Id));
            }
        }

        $this->render('update',array(
            'model'=>$model,
            'model2' => $model2,
        ));
    }

Answer 2

只需从控制器传递两个模型：

public function actionUpdate($id)
{
    $model=$this->loadModel($id);

    // assuming second model has same ID. Modify $id accordingly
    $model2=HotelRoomsChecked::model()->findByPk($id);

    if(isset($_POST['HotelRooms']) && isset($_POST['HotelRoomsChecked'])) //!
    {
        $model->attributes=$_POST['HotelRooms'];
        $model2->attributes=$_POST['HotelRoomsChecked']; //!

        if($model->save() && $model2->save()) //!
            $this->redirect(array('hotels/index','id'=>$model->Id));
    }

    $this->render('update',array(
        'model'=>$model,
        'model2'=>$model2, //!
    ));
}

并且只需按照您在问题中发布的方式使用表单中的第二个模型。