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I have the following code for updating a Yii model:

public function actionSettings($id) {
    if (!isset($_POST['save_hostname']) && isset($_POST['Camera']) && isset($_POST['Camera']['hostname'])) {
        $_POST['Camera']['hostname'] = '';
    }
    $model = $this->loadModel($id);
    $model->setScenario('frontend');
    $this->performAjaxValidation($model);
    if (isset($_POST['Camera'])) {
        $model->attributes = $_POST['Camera'];
        unset($model->api_password);
        if ($model->save()) {
            Yii::app()->user->setFlash('success', "Camera settings has been saved!");
        } else {
            Yii::app()->user->setFlash('error', "Unable to save camera settings!");
        }
    }
    $this->render('settings', array(
        'model' => $model,
    ));
}

This works fine, except in my model I have code like this:

<h1>Settings For: <?php echo CHtml::encode($model->name); ?></h1>

The problem is that, even when the user input fails validation, the h1 tag is having bad input echoed out into it. If the input fails the validation, the h1 attribute should stay the same.

I can 'reset' the $model variable to what is in the database before the view is returned, but this then means I don't get any error feedback / validation failed messages.

Is my only option to have 2 $models ($model and $data perhaps), one used for handling the form and the other for sending data to the page? Or does someone have a more elegant solution?

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1 回答 1

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performAjaxValidation 将所有保存属性分配给模型，因此这种行为是正常的。如果保存失败，我会重新加载模型。

$model->refresh();
于 2013-10-03T08:09:27.950 回答