How to modify two databes tables in one form in Yii

Question

I have created one form to eddit one table(hotel_rooms) in my database from my website page, it works perdectcly. But I need to include some fields,not all the table, from another table (hotel_rooms_checked)in the edditing web page of the privious table (hotel_rooms) how can I do it please ? where hotel_rooms.Id = hotel_rooms_checked.Id here is the code for the form

    <?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 -->

and this is the controller code

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,
        ));
    }

the view code is

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

the structure oh the tables

Hotel_Rooms
Id
HotelName
HotelRoomsType
.....

Hotel_Rooms_Checked
Id
DateCheckedIn
DateCheckedOut
....

Answer 1

Well, the question seems to be not clear enough. I would try to guess what you want though.

For the view, just add the fields for the other model (I suppose you have created a model for hotel_rooms_checked)

<?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 -->

And now for the controller, you could perform the validation for both models before actually saving.

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

Simply pass both models from the controller:

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, //!
    ));
}

And simply use the second model in your form the same way that you have posted in your question.