Rest ActiveController Yii2中id为的自定义操作

正在尝试在Rest API中实现GET方法，以查询用户状态，例如GET user/：id/status

因此，获取用户id#1的状态将调用/user/1/status

在配置中，我有：

'urlManager'   => [
'enablePrettyUrl'     => true,
'enableStrictParsing' => true,
'showScriptName'      => false,
'rules'               => [
    [
        'class'      => 'yii\rest\UrlRule',
        'controller' => [
            'v1/user'
        ],
        'extraPatterns' => [
            'GET status' => 'status',
        ],
        'tokens'     => [
            '{id}' => '<id:\\w+>'
        ],
    ],
],

用户控制器：

namespace api\modules\v1\controllers;

use yii\rest\ActiveController;

class UserController extends ActiveController {

    public $modelClass = 'api\modules\v1\models\User';

    public function actions() {
        $actions = parent::actions();

        unset(
            $actions[ 'index' ],
            $actions[ 'view' ],
            $actions[ 'create' ],
            $actions[ 'update' ],
            $actions[ 'delete' ],
            $actions[ 'options' ]
        );


        return $actions;
    }

    protected function verbs() {
        return [
            'status' => [ 'GET' ],
        ];
    }

    public function actionStatus( $id ) {

        return 1;
    }

}

---

但是现在我不确定如何实际返回调用的数据。

Wat你的意思是我不确定如何实际返回数据？我只是在actionStatus内部查询并返回JSON吗？不确定actionStatus中应该进行哪些操作

$user = User::findOne($id);
if ($user)
    return Json::encode(['success'=>true, 'data'=>$user->status]);
else
    return Json::encode(['success'=>false, 'message'=>"Can't find user"]);

$user = User::findOne($id);
if ($user)
    return Json::encode(['success'=>true, 'data'=>$user->status]);
else
    return Json::encode(['success'=>false, 'message'=>"Can't find user"]);