如何在yii中以json格式(application / json)获得响应?

时间:2010-05-13 06:28:19

标签: php yii

如何在yii中以json格式(application / json)获得响应?

8 个答案:

答案 0 :(得分:84)

对于Yii 1:

在(基础)控制器中创建此功能:

/**
 * Return data to browser as JSON and end application.
 * @param array $data
 */
protected function renderJSON($data)
{
    header('Content-type: application/json');
    echo CJSON::encode($data);

    foreach (Yii::app()->log->routes as $route) {
        if($route instanceof CWebLogRoute) {
            $route->enabled = false; // disable any weblogroutes
        }
    }
    Yii::app()->end();
}

然后只需在行动结束时致电:

$this->renderJSON($yourData);

对于Yii 2:

Yii 2 has this functionality built-in,在控制器操作结束时使用以下代码:

Yii::$app->response->format = \yii\web\Response::FORMAT_JSON;
return $data;

答案 1 :(得分:18)

$this->layout=false;
header('Content-type: application/json');
echo CJavaScript::jsonEncode($arr);
Yii::app()->end(); 

答案 2 :(得分:14)

对于控制器内的Yii2:

public function actionSomeAjax() {
    $returnData = ['someData' => 'I am data', 'someAnotherData' => 'I am another data'];

    $response = Yii::$app->response;
    $response->format = \yii\web\Response::FORMAT_JSON;
    $response->data = $returnData;

    return $response;
}

答案 3 :(得分:9)

$this->layout=false;
header('Content-type: application/json');
echo json_encode($arr);
Yii::app()->end(); 

答案 4 :(得分:5)

class JsonController extends CController {

    protected $jsonData;

    protected function beforeAction($action) {
        ob_clean(); // clear output buffer to avoid rendering anything else
        header('Content-type: application/json'); // set content type header as json
        return parent::beforeAction($action);
    }

    protected function afterAction($action) {
        parent::afterAction($action);
        exit(json_encode($this->jsonData)); // exit with rendering json data
    }

}

class ApiController extends JsonController {

    public function actionIndex() {
        $this->jsonData = array('test');
    }

}

答案 5 :(得分:0)

使用

的一种更简单的方法
echo CJSON::encode($result);

示例代码:

public function actionSearch(){
    if (Yii::app()->request->isAjaxRequest && isset($_POST['term'])) {
            $models = Model::model()->searchNames($_POST['term']);
            $result = array();
            foreach($models as $m){
                $result[] = array(
                        'name' => $m->name,
                        'id' => $m->id,
                );


            }
            echo CJSON::encode($result);
        }
}
欢呼:)

答案 6 :(得分:0)

在要呈现JSON数据的控制器操作中,例如:actionJson()

public function actionJson(){
    $this->layout=false;
    header('Content-type: application/json');
    echo CJSON::encode($data);
    Yii::app()->end(); // equal to die() or exit() function
}

查看更多Yii API

答案 7 :(得分:-1)

Yii::app()->end()

我认为这个解决方案不是结束应用程序流的最佳方式,因为它使用PHP的exit()函数,这意味着立即退出执行流程。是的,有Yii的onEndRequest处理程序和PHP的register_shutdown_function,但它仍然过于宿命。

对我来说,更好的方法就是这个

public function run($actionID) 
{
    try
    {
        return parent::run($actionID);
    }
    catch(FinishOutputException $e)
    {
        return;
    }
}

public function actionHello()
{
    $this->layout=false;
    header('Content-type: application/json');
    echo CJavaScript::jsonEncode($arr);
    throw new FinishOutputException;
}

因此,即使在之后,应用程序流仍继续执行。