Php 为数据对象laravel api添加更多值

如何将$st转换为数据对象我尝试了'st'=>$st，但对我不起作用请提前向我表示感谢

如何使用$st nowst:[{st:something}]需要响应类似的响应return response:：json[success=>true，message=>User found，st=>st，data=>Auth:：User->load['ch'=>function$query使用$st{$query->noGu->with'sp'；}，'sp'=>function$query{return Auth:：user->sp；}，'st'=>$st]]，200；编写更干净的代码怎么样？您可以在这个函数中使用$st，在这里您已经将它包含在use$st中，就像在任何其他地方一样…echo$st；

$st = 'Something';
return Response::json([
            "success" => true,
            "message" => "User found",
            "st" => $st,
            "data"    => Auth::user()->load(['ch' => function ($query) {
                            $query->noGu()
                                        ->with('sp');
                    },
                    'sp' => function($query){
                        return Auth::user()->sp;
                    }])
        ], 200);

        $st = 'Something';
    return Response::json([
                "success" => true,
                "message" => "User found",
                "st" => $st,
                 //Take the use statement
                "data"    => Auth::user()->load(['ch' => function ($query) use ($st) {
                                //Now you can use $st inside this function
                                $query->noGu()->with('sp');
                        },
                        'sp' => function($query){
                            return Auth::user()->sp;
                        }])
            ], 200);