dongzhan5286
2016-12-06 06:01
浏览 21
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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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我有以下代码,我想使用名为link而不是id的字段查找模型。 但是,它似乎没有产生任何结果。 我哪里可以弄错? 它返回404

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

注意:在搜索模型中,我尝试添加此项:

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

  • duangekui7451 2016-12-08 06:35
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    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>',
            ),
            ],
        ],
    
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  • dongmeng1875 2016-12-06 06:57

    404 means your action was not found by url manager. link parameter must be in the url (http://yourapp/controller/view?link=something) because you definded it as actionView($link)

    Requesting http://yourapp/controller/view will give you 404 error.

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  • dsla94915 2016-12-06 07:33

    If you want a single model you need one() and not all()

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

    With one() method you retrive just a model and in your $model you have the data you need ..

    If you sue all() lie you did you retrive a collection of models and for accessing a single model you must set a proper index eg:

    $my_model = $model[0];
    
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  • drnf09037160 2016-12-06 11:02

    In UrlManager config you propably have:

    '<controller>/<action>/<id:\d+>' => '<controller>/<action>',
    

    It means default rewrite rules use id as param and it has to be digit, so you can't use controller/view/link. Just to be sure, change action name from actionView to actionTest and then call URL controller/test?link=linklink.

    Other solution is to use URL like this: controller/view?link=linklink

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  • doushan1863 2016-12-06 12:52

    The problem is in your url

    case 1)

    http://localhost/projectName/backend/web/controllerName/actionName/username/rushil

    Gives output : Not Found (#404)

    case 2)

    http://localhost/projectName/backend/web/controllerName/actionName/rushil

    Gives output : Not Found (#404)

    case 3)

    http://localhost/projectName/backend/web/controllerName/actionName?username=rushil

    this will work.

    Solution : check your url and pass link parameter in url as shown in case 3

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