Question

我正在使用laravel中的联接查询从数据库中获取数据并传递给json，并在数组中获得一些结果，但是我想要下面给出的对象

控制器代码：

$resultPastActivity= DB::table('table_user_create_activity')
                            ->join('table_sub_category','table_user_create_activity.selected_activity_id', '=', 'table_sub_category.sub_category_id')
                            ->select('sub_category_name','area','activity_type','activity_date','start_time','end_time')
                            ->whereDate('activity_date', '<', $todayDate)
                            ->where('user_id',$user_id)
                            ->get();

 return response()->json(['success' => '1','data' =>$resultPastActivity]);

上面的代码将给出以下数组中的json实际上我想要对象中的json

{
    "success": "1",
    "data": [
        {
            "sub_category_name": "Badminton",
            "area": "Rankala lake",
            "activity_type": "1",
            "activity_date": "2018-01-12",
            "start_time": "15:04:49",
            "end_time": "20:05:69"
        },
        {
            "sub_category_name": "Football",
            "area": "Devakar panad",
            "activity_type": "1",
            "activity_date": "2018-01-15",
            "start_time": "15:04:49",
            "end_time": "20:05:69"
        },
    ]
}



i want json as follows
{
    "success": "1",
    "data": {
        {
            "sub_category_name": "Badminton",
            "area": "Rankala lake",
            "activity_type": "1",
            "activity_date": "2018-01-12",
            "start_time": "15:04:49",
            "end_time": "20:05:69"
        },
        {
            "sub_category_name": "Football",
            "area": "Devakar panad",
            "activity_type": "1",
            "activity_date": "2018-01-15",
            "start_time": "15:04:49",
            "end_time": "20:05:69"
        },
    }
}

Answer 1

尝试一下，json_encode（$ json，JSON_FORCE_OBJECT）JSON_FORCE_OBJECT

$result=array();
$result["0"]=$resultPastActivity;
$json=json_encode((object)$result,JSON_FORCE_OBJECT);
return response()->json(['success' => '1','data' =>$json]);

Answer 2

您想要的不是有效的JSON，您可以在此处进行检查：https://jsonformatter.curiousconcept.com/

因此，您无法创建这样的输出，并且如果强制执行该输出，则将收到该输出的应用程序将无法正确解析它，因此这没有任何意义。

在laravel 5.7中将json数组转换为json对象

2 个答案: