如何迭代Yii CActiveDataProvider对象？

Question

如何迭代数据提供程序对象？我希望访问返回的每一行的“名称”字段，并构建一个列表。你能帮上忙吗？

表/模型categories的表结构

CREATE TABLE IF NOT EXISTS `categories` (
  `idCategory` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(64) NOT NULL,
  PRIMARY KEY (`idCategory`) USING BTREE
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=55 ;

*我的控制器Categories*中的函数

    $names = array();
    public function returnCategoryNames()
{
    $dataProvider= new CActiveDataProvider('Categories');
    $dataProvider->setPagination(false);
    $count = $dataProvider->totalItemCount();

    for($i = 0; $i < $count; $i++){

             // this is where I am lost...
             $myname = $dataProvider->data[$i]->name;
             array_push($names, $myname);

    }   

       return $names;

}

Answer 1

试试这个：

public function returnCategoryNames()
{
  $dataProvider= new CActiveDataProvider('Categories');
  $dataProvider->setPagination(false);
  //$count = $dataProvider->totalItemCount();
  $names = array();
  foreach($dataProvider->getData() as $record) {
    $names[] = $record->name;
  }
  return array_unique($names);
}

但是，您不需要使用数据提供程序，而只需使用模型

foreach(Categories::model()->findAll() as $record) {

Answer 2

在Yii2中，这种情况也发生了变化：

    $searchModel = new ModelSearch();
    $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

    foreach($dataProvider->getModels() as $record) {
        echo $record->id . '<br>';
    }

    exit();

参考资料：getModels()

Answer 3

使用@ben的解决方案，您将同时查询所有行。你可能会有记忆问题。

使用以下解决方案，您将从10乘10(默认的CPagination.pageSize值)获取类别：

        $dataProvider = new CActiveDataProvider('Categories', array(
            'pagination' => array(
                'validateCurrentPage' => false
            ),
        ));
        $pagination = $dataProvider->pagination;
        while ($categories = $dataProvider->getData(true)){
            foreach ($categories as $category) {
                //...
            }
            $pagination->currentPage++;
        }