Question

我想接收一个数据数组，像这样：

    "data": [
    "6": {
        "title": "..",
        "photo": ..
    }
]

但是我正在接收一个对象：

    "data": {
    "6": {
        "title": "..",
        "photo": ".."
    }
}

另一个有趣的事实是，对于第一页，我正在收到需要的内容：

        "data": [
    "1": {
        "title": "..",
        "photo": ..
    },
.
.
.
        "6": {
            "title": "..",
            "photo": ..
        }
]

这是我在控制器中的功能：

        $lists = [];
    $cooks = cook::all();
    foreach ($cooks as $cook) {
        $list = [
            'title' => $cook->title,
            'photo' => $cook->photos->path,
            'recipe' => $cook->recipe->description
        ];
        $lists[] = $list;
    }
    $collection = $this->paginate($lists, $perPage = 6, $page = null, $options = []);
    return new mainCollection($collection);

我具有以下分页功能：

    public function paginate($items, $perPage = 15, $page = null, $options = [])
{
    $page = $page ?: (Paginator::resolveCurrentPage() ?: 1);
    $items = $items instanceof Collection ? $items : Collection::make($items);
    return new LengthAwarePaginator($items->forPage($page, $perPage), $items->count(), $perPage, $page, $options);
}

以下是我的resourceColection（mainColection）：

public function toArray($request)
{
    return [
        'data' => $this->collection,
        'meta' => ['pagination' => $this->pagination]
    ];
}

Answer 1

由于正在使用集合，因此可以使用laravel映射。

$array =  $this->collection->map(function ($item, $key) {
    return [
       'title': "tort",
        ...
    ]
});

并将其编码为json

 public function toArray($request)
   {
    return [
        'data' => json_encode($array),
        'meta' => ['pagination' => $this->pagination]
    ];
   }

您还可以在此处阅读文档：https://laravel.com/docs/5.8/collections

如何使用数组而不是对象来响应数据

1 个答案: