Question

我尝试使用Laravel's API Resources来处理某些JSON，并且我无法有条件地加载关系，尽管它已被迫加载。

我的控制器：

$games = Game::with('availableInterests')->get();

在我看来，我是json编码要在VueJS中使用的集合

games = @json(new \App\Http\Resources\GameCollection($games)),

GameCollection - 与Laravel为我生成的类保持不变。

<?php

namespace App\Http\Resources;

use Illuminate\Http\Resources\Json\ResourceCollection;

class GameCollection extends ResourceCollection
{
    /**
     * Transform the resource collection into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
        return parent::toArray($request);
    }
}

GameResource

class GameResource extends JsonResource
{
    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
        return [
            'id' => $this->id,
            'name' => $this->name,
            'thumbnail_url' => $this->thumbnail_url,
            'available_interests' => Interest::collection($this->whenLoaded('availableInterests')),
        ];
    }
}

游戏模型的关系

public function availableInterests() : BelongsToMany
{
    return $this->belongsToMany(Interest::class);
}

我尝试将$this->whenLoaded('availableInterests')更改为$this->whenLoaded('available_interests')但没有运气。我没有运气就检查了我的拼写。

为什么这个conditional relationship出现在json中？

即使删除$this->whenLoaded()也不会显示此关系。

Answer 1

在这种情况下，我认为你不需要GameCollection。

我会尝试这样做：

InterestResource.php （创建新课程）

class InterestResource extends JsonResource
{
    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
        return parent::toArray($request);
        // or what ever array structure you want
    }
}

<强> GameResource.php

class GameResource extends JsonResource
{
    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
        return [
            'id' => $this->id,
            'name' => $this->name,
            'thumbnail_url' => $this->thumbnail_url,
            'available_interests' => InterestResource::collection($this->whenLoaded('availableInterests')),
        ];
    }
}

您的控制器

$games = Game::with('availableInterests')->get();

您的观点

games = @json(\App\Http\Resources\GameResource::collection($games)),

条件关系未出现在Laravel Resource API中

1 个答案: