dongzhan5286 2016-12-06 06:01
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如何在Yii 2中使用id以外的字段查找记录

I have the following code where I want to find a model using a field called link instead of id. However, it doesn't seem to produce any results. Where could I be getting it wrong? It returns 404

public function actionView($link)
    {
        $model = News::find()->where(['link'=>$link])->all();
        return $this->render('view', [
            'model' => $model,
        ]);
    }

NB: in the search model, I have tried adding this:

 $query->andFilterWhere([
            'id' => $this->id,
            'link'=>$this->link,
            'category' => $this->category,
            'date' => $this->date,
            'userid' => $this->userid,
            'featured' => $this->featured,
        ]);
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5条回答 默认 最新

  • duangekui7451 2016-12-08 06:35
    关注

    For anyone who may encounter such an error, the problem is after using pretty urls, you need to add rules to map the url format. C:\xampp\htdocs\your_project\frontend\config\main.php under urlManager add the rules like so:

    'urlManager' => [
            'class' => 'yii\web\UrlManager',
            // Disable index.php
            'showScriptName' => false,
            // Disable r= routes
            'enablePrettyUrl' => true,
            'rules' => array(
                    '<controller:\w+>/<id:\d+>' => 'news/view',
                 '<controller:\w+>/<link>' => 'news/latestnews', // This is required for $link parameter
                    '<controller:\w+>/<action:\w+>/<id:\d+>' => '<controller>/<action>',
                '<controller:\w+>/<action:\w+>/<link>' => 'news/latestnews',
                    '<controller:\w+>/<action:\w+>' => '<controller>/<action>',
            ),
            ],
        ],
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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