Question

我已经从我的网站页面创建了一个表格来编辑我的数据库中的一个表（hotel_rooms），它可以正常工作。但我需要在私人表（hotel_rooms）的edditing网页中的另一个表（hotel_rooms_checked）中包含一些字段，而不是所有表格。我怎么能这样做？其中hotel_rooms.Id = hotel_rooms_checked.Id 这是表格的代码

    <?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,
        ));
    }

视图代码是

<?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',
        ....
    ),
)); ?>

表格的结构

Hotel_Rooms
Id
HotelName
HotelRoomsType
.....

Hotel_Rooms_Checked
Id
DateCheckedIn
DateCheckedOut
....

Answer 1

嗯，问题似乎不够明确。我会尽量猜猜你想要什么。

对于view，只需为其他model添加字段（我猜您已为hotel_rooms_checked创建了模型）

<?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执行验证。

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, //!
    ));
}

只需在表单中使用第二个模型，就像在问题中发布一样。

如何在Yii中修改一个表单中的两个数据库表

2 个答案: