【发布时间】:2019-05-23 11:46:27
【问题描述】:
我正在使用 laravel 中的连接查询从数据库中获取数据并传递给 json 并在数组中得到一些结果,但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]);
上面的代码将给出数组中的以下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"
},
}
}
【问题讨论】:
-
你试过 flatten() 吗?这是集合的一种方法。
-
快速完成此操作
$obj = json_decode(json_encode($array)); -
我试过但得到相同的 json 和数组?建议另一种方法
-
这不是有效的 json,要在一个结构中有多个对象,您需要将其包装在一个数组中,您想要一个不可能的多个对象的对象
标签: laravel