Question

默认情况下，对于所有API，我都是从laravel分形库获取数据作为对象的。

我不想要数据，而是想要一些自定义名称而不是数据，但是我无法做到这一点。

如何自定义分形库中的代码以获取项目而不是数据。

Answer 1

创建DataSerializer类：

namespace App\{Your Project Name}\Serializers;

use League\Fractal\Serializer\ArraySerializer;

class DataSerializer extends ArraySerializer
{
    public function collection($resourceKey, array $data)
    {
        if ($resourceKey) {
            return [$resourceKey => $data];
        }

        return $data;
    }
    public function item($resourceKey, array $data)
    {
        if ($resourceKey) {
            return [$resourceKey => $data];
        }
        return $data;
    }
}

在基本控制器中添加setResponseData函数：

public function setResponseData($default=true, $data, $transformer, $includes = null){

    if($default){
        $resource = fractal($data, $transformer);

        if($includes){
            $resource->parseIncludes($includes);
        }
        return $resource;
    }

    $resource = null;
    if($data instanceof LengthAwarePaginator){
        $dataCollection = $data->getCollection();
        $resource = new Collection($dataCollection, $transformer, 'data');
        $resource->setPaginator(new IlluminatePaginatorAdapter($data));
    } elseif($data instanceof \Illuminate\Database\Eloquent\Collection){
        $resource = new Collection($data, $transformer, 'data');
    } elseif($data instanceof Model){
        $resource = new Item($data, $transformer, 'data');
    } else{
        return [];
    }

    $manager = new Manager();
    $manager->setSerializer(new DataSerializer());

    if($includes){
        $manager->parseIncludes($includes);
    }

    $content = [];
    if($resource){
        $content = $manager->createData($resource)->toArray();
    }

    return $content;
}

从派生控制器中调用setResponseData函数：

$response = $this->setResponseData(false, $data, new YourTransformer(), ['include-1']);

如何从laravel fractol库的响应中删除数据对象？

1 个答案: