如何将数据从DB表中提取为json格式？LARAVEL 5.8

Question

我是一名业余的前端开发人员，为我的大学做一个网络开发项目。我要单独做这个项目已经完成了网站的前端部分。我使用一个临时API来获取在我的视图上传递的数据。现在，由于我必须使用数据库，我构建了数据库，并试图将数据获取到我的控制器。这对我来说太复杂了，到目前为止几乎没有任何进展。现在我正在尝试获取控制器上的一个变量JSON数据，它包含特定行的所有表键和值。响应包含受保护的数据，我不知道如何以正确的方式获取它们。

如有任何建议或建议，将不胜感激。

我的测试服务器的控制器

<?php

namespace App\Http\Controllers;
use App\circuits;
use App\constructor_results;
use App\constructors;
use App\driver_standings;
use App\drivers;
use App\lap_times;
use App\pit_stops;
use App\qualifying;
use App\races;
use App\results;
use App\seasons;
use App\status;


use Illuminate\Http\Request;
use Illuminate\Support\Facades\DB;

class HomeController extends Controller
{
    /**
     * Create a new controller instance.
     *
     * @return void
     */
    
    public function __construct()
    {
        $this->middleware('auth');
    }

    /**
     * Show the application dashboard.
     *
     * @return \Illuminate\Contracts\Support\Renderable
     */
    public function index()
    {

        $data = DB::table('circuits')
        ->select('circuits.*')
        ->where(['circuitId' => '6'])
        ->get();

        print_r(response()->json($data));       

    }
 
}

响应(想要的值显示在行尾)

Illuminate\Http\JsonResponse Object ( [data:protected] => [{"circuitId":"6","circuitRef":"monaco","name":"Circuit de Monaco","location":"Monte-Carlo","country":"Monaco","lat":"43.7347","lng":"7.42056","alt":"7","url":"http:\/\/en.wikipedia.org\/wiki\/Circuit_de_Monaco"}] [callback:protected] => [encodingOptions:protected] => 0 [headers] => Symfony\Component\HttpFoundation\ResponseHeaderBag Object ( [computedCacheControl:protected] => Array ( [no-cache] => 1 [private] => 1 ) [cookies:protected] => Array ( ) [headerNames:protected] => Array ( [cache-control] => Cache-Control [date] => Date [content-type] => Content-Type ) [headers:protected] => Array ( [cache-control] => Array ( [0] => no-cache, private ) [date] => Array ( [0] => Tue, 09 Nov 2021 18:09:00 GMT ) [content-type] => Array ( [0] => application/json ) ) [cacheControl:protected] => Array ( ) ) [content:protected] => [{"circuitId":"6","circuitRef":"monaco","name":"Circuit de Monaco","location":"Monte-Carlo","country":"Monaco","lat":"43.7347","lng":"7.42056","alt":"7","url":"http:\/\/en.wikipedia.org\/wiki\/Circuit_de_Monaco"}] [version:protected] => 1.0 [statusCode:protected] => 200 [statusText:protected] => OK [charset:protected] => [original] => Illuminate\Support\Collection Object ( [items:protected] => Array ( [0] => stdClass Object ( [circuitId] => 6 [circuitRef] => monaco [name] => Circuit de Monaco [location] => Monte-Carlo [country] => Monaco [lat] => 43.7347 [lng] => 7.42056 [alt] => 7 [url] => http://en.wikipedia.org/wiki/Circuit_de_Monaco ) ) ) [exception] => )

Answer 1

看看https://laravel.com/docs/8.x/eloquent-resources

(它应该适用于5.8以及https://laravel.com/docs/5.8/eloquent-resources，其想法是一样的)

当然，您可以使用

方法将有说服力的模型或集合转换为JSON；但是，雄辩的资源为您的模型及其关系的JSON序列化提供了更细粒度和更健壮的控制。

因此，您可以添加一个新的电路雄辩模型，描述所有字段等等，并添加toJson方法，它将返回数组(在那里，您可以映射您希望从模型返回的内容)。

我更喜欢使用API参考资料，因为在特性中，您可能需要不同地方的不同响应。

<?php
namespace App\Http\Resources;

use Illuminate\Http\Resources\Json\JsonResource;

class CircuitResource extends JsonResource
{
    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
        //add all fields which you want to return from the Circuit model
        return [
            'id' => $this->id,
            'created_at' => $this->created_at,
            'updated_at' => $this->updated_at,
        ];
    }
}

然后在控制器中，您将使用它就像

use App\Http\Resources\CircuitResource;
use App\Models\Circuit;

...

return response()->json(new CircuitResource(Circuit::find(6)));

模型将是

<?php
namespace App\Models;

use Illuminate\Database\Eloquent\Model;

class Circuit extends Model
{
    /**
     * The primary key associated with the table.
     *
     * @var string
     */
    protected $primaryKey = 'circuitId';
}

要找到1项，最好使用find()方法。但是，如果搜索多个，则会收到Collection对象。在这种情况下，您将需要在创建的资源类上使用适当的方法，并且应该从操作中返回

return response()->json(new CircuitResource::collection(Circuit::all()));