Yii框架中出现未知错误


Unknown Error in Yii Framework

我的Yii音乐商店应用程序中有这段代码,但我不知道错误。我的想法是显示/浏览专辑、艺术家和流派的类别/标准。但它总是显示"其他条件"。我已尝试修复它,但找不到错误。代码如下。希望任何人都能修复它。谢谢所有

<?php
class StoreController extends Controller
{
public $message;
public function actionIndex()
{
    $this->message = "Hello from Store.Index()";
    $this->render('index', array('content'=>$this->message));
}
public function actionBrowse()
{
    if(isset($_GET["gid"])){
        $gid = $_GET["gid"];
        $genreCriteria = new CDbCriteria();
        $genreCriteria->select = "`GenreId`, `Name`, `Description`";
        $genreCriteria->condition = "genreId = " . $_GET["gid"];
        $artistCriteria = new CDbCriteria();
        $artistCriteria->alias = "t1";
        $artistCriteria->select = "DISTINCT `t1`.`Name`, `t1`.`ArtistId`";
        $artistCriteria->join = "LEFT JOIN `tbl_album` ON `tbl_album`.`ArtistId` = `t1` . `ArtistId`";
        $artistCriteria->order = "`t1`.`ArtistId` ASC";
        $albumCriteria = new CDbCriteria();
        $albumCriteria->alias = "t2";
        $albumCriteria->select = "`AlbumId`, `GenreId`, `Title`, `Price`, `AlbumArtUrl`";
        $albumCriteria->condition = "`GenreId` = " . $_GET["gid"];
        $albumCriteria->order = "`ArtistId` ASC";
        $this->render('index', array('Albums' => Album::model()->findAll($albumCriteria),
                                     'Artist' => Artist::model()->findAll($artistCriteria),
                                     'Genre' => Genre::model()->findAll($artistCriteria)));
    }else{
        $this->message = "Hello from Store.Browse()";
        $this->render('index', array('content'=>$this->message));
    }
}
public function actionDetails()
{
    $this->message = "Hello from Store.Details()";
    $this->render('index', array('content'=>$this->message));
}
// Uncomment the following methods and override them if needed
/*
public function filters()
{
    // return the filter configuration for this controller, e.g.:
    return array(
        'inlineFilterName',
        array(
            'class'=>'path.to.FilterClass',
            'propertyName'=>'propertyValue',
        ),
    );
}
public function actions()
{
    // return external action classes, e.g.:
    return array(
        'action1'=>'path.to.ActionClass',
        'action2'=>array(
            'class'=>'path.to.AnotherActionClass',
            'propertyName'=>'propertyValue',
        ),
    );
}
*/
}

我确信答案很简单,您的页面URL缺少"gid"及其值。这是"如果isset"条件属于"其他"部分的唯一合理解释。

你可以做一个快速的确认测试。转到浏览器,输入www.yoursitename.com/store/browse?gid=111,然后看看它是否能得到您需要的结果。请记住,我不知道你的确切URL及其结构,所以这只是一个例子,但"/浏览?gid=111"部分非常重要。它简单地为$_POST数组创建"gid",并将其值设置为"111"。

附加说明:与"isset"一起,您还可以检查"gid"值是否为空,有时这种方式会很有用,也更安全。

此外,这是我的Yii代码版本:

$genreCriteria = new CDbCriteria();
    $genreCriteria->select = "GenreId, Name, Description";
    $genreCriteria->condition = "genreId = " . $_GET["gid"];
    $artistCriteria = new CDbCriteria();
    $artistCriteria->alias = "t1";
    $artistCriteria->select = "DISTINCT t1.Name, t1.ArtistId";
    $artistCriteria->join = "LEFT JOIN tbl_album ON tbl_album.ArtistId = t1.ArtistId";
    $artistCriteria->order = "t1.ArtistId ASC";
    $albumCriteria = new CDbCriteria();
    $albumCriteria->alias = "t2";
    $albumCriteria->select = "AlbumId, GenreI, Title, Price, AlbumArtUrl";
    $albumCriteria->condition = "GenreId = " . $_GET["gid"];
    $albumCriteria->order = "ArtistId ASC";

正如你所看到的,我只是去掉了引号。这段代码没有在Yii中进行本地测试,但查询本身对我来说很好