Question

我的Yii音乐商店应用程序中有此代码，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”。我试图修复它，但我找不到错误。代码如下。希望有人能解决它。谢谢大家

<?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',
        ),
    );
}
*/
}

Answer 1

我很确定答案非常简单，您的网页网址缺少“gid”及其价值。这是“if isset”条件属于“else”部分的唯一合理解释。

您可以进行快速确认测试。转到您的浏览器，输入 www.yoursitename.com/store/browse?gid=111 ，然后查看是否能获得所需的结果。请注意，我不知道您的确切网址及其结构，所以这只是一个示例，但“/ browse？gid = 111 ”这一部分非常重要。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，有时候这样有用且更安全。

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";

如您所见，我只是删除了引号。这个代码没有在Yii本地测试，但查询本身 - 对我来说很好。

Yii框架中的未知错误

1 个答案: