我有这段代码
public function actionIndex() {
$searchModel = new DraftSearch();
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
return $this->render('index', [
'dataProvider' => $dataProvider,
]);
}
我想以JSON格式返回$ dataprovider,请知道如何在Yii2中执行此操作,谢谢
答案 0 :(得分:0)
好像你需要\Yii::$app->response->format = \yii\web\Response::FORMAT_JSON
如果你需要json中的返回值,你应该使用
public function actionIndex()
{
\Yii::$app->response->format = \yii\web\Response::FORMAT_JSON;
$searchModel = new DraftSearch();
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
return $dataProvider;
}
或者,如果你需要渲染中的dataProvider,你可以试试json_encode
public function actionIndex() {
$searchModel = new DraftSearch();
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
$myJsonDataProvider json_encode($dataProvider);
return $this->render('index', [
'dataProvider' => $myJsonDataProvider,
]);
}