yii2 rest api tuning

Question

我正在尝试使用yii2的REST api获取json格式中所有项目的列表。基本上，它返回一个没有排序的20个项目的列表。但我需要某种东西。如果我不得不告诉他们，我可能会重新定义控制器的动作指数这样的方式：

public function actionIndex() 
{ 
    $modelClass = $this->modelClass; 
    $dictionaries = $modelClass::find()->select([ 
      '{{site_search_dictionary}}.*', // select all customer fields 
      'COUNT(*) AS dd' // calculate orders count 
     ]) 
     ->joinWith('queries') // ensure table junction 
     ->groupBy('dictionary_id') // group the result to ensure aggregation function works 
      ->orderBy('dictionary')->all(); 

    return $this->render('index', [ 
     'dictionaries' => $dictionaries, 
    ]); 
} 

但我能做些什么来改变同样的方式回应休息查询？

Answer 1

我想你可以用它舔这个

public function actionIndex() 
{ 
$modelClass = $this->modelClass; 
$dictionaries = $modelClass::find()->select([ 
     '{{site_search_dictionary}}.*', // select all customer fields 
     'COUNT(*) AS dd' // calculate orders count 
    ]) 
    ->joinWith('queries') // ensure table junction 
    ->groupBy('dictionary_id') // group the result to ensure aggregation function works 
     ->orderBy(['dictionary_id'=>SORT_ASC])->all(); 

return $this->render('index', [ 
    'dictionaries' => $dictionaries, 
]); 
} 

Answer 2

你可以重新定义prepareDataProvider索引操作。

public function actions() 
     { 
      $actions = parent::actions(); 
      $actions['index']['prepareDataProvider'] = [$this, 'prepareDataProvider']; 

      return $actions; 
     } 

这将从您的控制器调用prepareDataProvider方法而不是内置方法。

例如：

public function prepareDataProvider() 
     { 
      $searchModel = new SiteDictionarySearch(); 
      $dataProvider = $searchModel->search(\Yii::$app->request->queryParams); 
      return $dataProvider; 
     } 

你应该让你的模型，这样的需求的搜索模式。要返回所有项目更好地利用ArrayDataProvider与find()->asArray()

这里是回报应该什么样子：

return new ArrayDataProvider([ 
       'allModels' => $rows, 
       'pagination' => [ 
        'pageSize' => count($rows), 
       ], 
      ]);