Question

在_form.php文件中我有来自两个不同模型的文本框，我已对create控制器进行了更改以插入两个模型的记录，但问题出在update控制器中，如何加载第二个模型的文本框中的第二个模型的数据？
只会填充与第一个模型相关的文本框。

Answer 1

只需将两个模型渲染到控制器中的update视图即可。

$this->render('update', array(
    'model1' => $model1,
    'model2' => $model2,
));

并在_form.php中调用类似这样的文本框

<?php echo $form->labelEx($model1, ‘data1’); ?>
<?php echo $form->textField($model1, ‘data1’, array(‘size’ => 60, ‘maxlength’ => 250)); ?>

<?php echo $form->labelEx($model2, ‘data2’); ?>
<?php echo $form->textField($model2, ‘data2’, array(‘size’ => 60, ‘maxlength’ => 250)); ?>

希望这会有所帮助。

修改

由于您可能正在为create视图使用相同的_form.php页面，因此您需要创建另一个_form.php文件，例如 _formUpdate.php [_form.php的副本]页面并从您的 update.php 调用渲染_formUpdate.php而不是_form.php并进行上述更改

Answer 2

_form

    <div class="form">

<?php $form=$this->beginWidget('CActiveForm', array(
    'id'=>'tabel1-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,'name'); ?>
        <?php echo $form->textField($model,'name',array('size'=>44,'maxlength'=>44)); ?>
        <?php echo $form->error($model,'name'); ?>
    </div>
    <div class="row">
        <?php echo $form->labelEx(Table2::model(),'address'); ?>
        <?php echo $form->textField(Table2::model(),'address',array('size'=>44,'maxlength'=>44)); ?>
        <?php echo $form->error(Table2::model(),'address'); ?>
    </div>

    <div class="row buttons">
        <?php echo CHtml::submitButton($model->isNewRecord ? 'Create' : 'Save'); ?>
    </div>

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

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

update.php

    <?php
$this->breadcrumbs=array(
    'Tabel1s'=>array('index'),
    $model->name=>array('view','id'=>$model->id),
    'Update',
);

$this->menu=array(
    array('label'=>'List Tabel1', 'url'=>array('index')),
    array('label'=>'Create Tabel1', 'url'=>array('create')),
    array('label'=>'View Tabel1', 'url'=>array('view', 'id'=>$model->id)),
    array('label'=>'Manage Tabel1', 'url'=>array('admin')),
);
?>

<h1>Update Tabel1 <?php echo $model->id; ?></h1>

<?php echo $this->renderPartial('_formUpdate', array('model'=>$model,'model2'=>$model2)); ?>

Tabel1Controller.php

 public function actionCreate()
    {
        $model=new Tabel1;
    $model2=new Table2;
        // Uncomment the following line if AJAX validation is needed
        // $this->performAjaxValidation($model);

        if(isset($_POST['Tabel1'])&&isset($_POST['Table2']))
        {
            $model->attributes=$_POST['Tabel1'];
                        $model2->attributes=$_POST['Table2'];
            if($model->save()&&$model2->save())
                $this->redirect(array('view','id'=>$model->id));
        }

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

    /**
     * Updates a particular model.
     * If update is successful, the browser will be redirected to the 'view' page.
     * @param integer $id the ID of the model to be updated
     */
    public function actionUpdate($id)
    {
        $model=$this->loadModel($id);
    $model2=$this->loadModel2($id);
        // Uncomment the following line if AJAX validation is needed
        // $this->performAjaxValidation($model);
//echo $model2->address;
        if(isset($_POST['Tabel1'])&&isset($_POST['Table2']))
        {
            $model->attributes=$_POST['Tabel1'];
                        $model2->attributes=$_POST['Table2'];
            if($model->save()&&$model2->save())
                $this->redirect(array('view','id'=>$model->id));
        }

        $this->render('update',array(
            'model'=>$model,
                    'model2'=>$model2
        ));
    }
 public function loadModel2($id)
    {
        $model=Table2::model()->findByPk(1);
        if($model===null)
            throw new CHttpException(404,'The requested page does not exist.');
        return $model;
    }

如何在更新控制器中加载多个模型

2 个答案: