Question

我正在为front end and mobile apps实施API，现在我正在search function工作，用户可以在其中输入column name或任何data我应该能够以JSON格式提供他们请求的数据

所以我做了这个

在我的控制器中

public function getSearchResults(Request $request) {

    $data = $request->get('data');

    $search_drivers = Driver::where('agent_id', 'like', "%{$data}%")
                     ->orWhere('registration_center', 'like', "%{$data}%")
                     ->orWhere('registration_id', 'like', "%{$data}%")
                     ->orWhere('sponsor_name', 'like', "%{$data}%")
                     ->orWhere('event_name', 'like', "%{$data}%")
                     ->orWhere('registration_id', 'like', "%{$data}%")
                     ->orWhere('profile_photo', 'like', "%{$data}%")
                     ->orWhere('first_name', 'like', "%{$data}%")
                     ->orWhere('last_name', 'like', "%{$data}%")
                     ->get();

    return Response::json([
        'data' => $search_drivers
    ]);     
}

使用参数i搜索获取JSON响应

http://localhost:8000/api/v1/search?data=center 

//Response
{
    "data": [
        {
            "id": 1,
            "agent_id": "201701",
            "registration_center": "Center",
            "registration_date": "12-01-2017",
            "sponsor_name": "Sponser Name",
            "event_name": "Event Name",
            "registration_id": "45345343543353",
            "profile_photo": "",
            "first_name": "",
            "last_name": "",
          }
    ]
}

但我如何过滤像这样的json

http://localhost:8000/api/v1/search?data=center 

//Response
{
    "data": [
        {
            "registration_center": "center",
          }
    ]
}

是否可以这样做，请您告诉我或给我一些示例如何query and return as json

谢谢

Answer 1

将['registration_center']添加到get()：

->get(['registration_center']);

或者在select()之前使用get()：

->select('registration_center')
->get();

Laravel 5.3过滤API

1 个答案: