Php 我想用Yii显示最后4页?
学习编程Yii,我想显示最后4页: SiteController.phpPhp 我想用Yii显示最后4页?,php,yii,Php,Yii,学习编程Yii,我想显示最后4页: SiteController.php public function actionStart() { $featured = Page::model()->findAllByAttributes( array(), $condition = 'featured = :featureId', $params = array(
public function actionStart()
{
$featured = Page::model()->findAllByAttributes(
array(),
$condition = 'featured = :featureId',
$params = array(
':featureId' => 1,
)
);
$this->render('/layouts/start/start', array('featured'=>$featured));
}
<?php print_r($this->featured); ?>
/layouts/start/start.php
public function actionStart()
{
$featured = Page::model()->findAllByAttributes(
array(),
$condition = 'featured = :featureId',
$params = array(
':featureId' => 1,
)
);
$this->render('/layouts/start/start', array('featured'=>$featured));
}
<?php print_r($this->featured); ?>
后一个文件不显示任何内容,应该是一个包含数据的数组,如何获取它?删除$this for featured
<?php echo print_r($featured, true); ?>
在这里,您将向视图发送值数组(关联数组)。您可以通过调用数组键
来访问这些值
因此,您的代码应该是
<?php echo print_r($featured); ?>
我正在发送我的姓名和年龄以供查看。为了显示我的姓名和年龄,我应该叫钥匙
echo $myName;
echo $myAge;
我认为您应该先阅读文档。不能将start.php用于SiteController。根据Yii规则,start.php应该驻留在站点的视图中。您必须使用cgridview或Yii内置的某种网格输出特征变量,这很容易,但正如@Namelus所说,阅读文档以了解如何执行