多对多关系的分页

Question

我的数据库中有products、categories 和category_product（数据透视表）。

我想在相同的响应中返回类别信息和产品与分页。

我知道我可以使用Product::with('categories')->pagination(20)，

但我不想将类别附加到每个产品。

让产品属于特定类别的正确方法是什么？

我已经尝试过了，但我无法获得分页：

$category = Category::where('slug', $slug)->first();
$products = $category->products()->paginate(20);
return response()->json([
    'category' => new CategoryResource($category),
    'products' => ProductResource::collection($products),
]);

这是我的产品资源

return [
    'id' => $this->id,
    'name' => $this->name,
    'description' => $this->description,
    'code' => $this->code,
    'image' => $this->image,
    'quantity' => $this->quantity,
    'price' => $this->price,
    'slug' => $this->slug,
    'sort_order' => $this->sort_order,
    'created_at' => $this->created_at,
    'updated_at' => $this->updated_at,
    'categories' => CategoryResource::collection($this->whenLoaded('categories')),
];

Answer 1

这看起来像是集合中的issue with the way data is returned。

最简单的解决方案是：

return response()->json([
    'category' => new CategoryResource($category),
    'products' => ProductResource::collection($products)->response()->getData(true),
]);

Answer 2

你可以尝试如下

$category = Category::where('slug', $slug)
            ->with('products', function($query) {
                $query->paginate(20);
            })
            ->first();

return response()->json([
    'category' => new CategoryResource($category),
    'products' => ProductResource::collection($category->products),
]);

希望这是您正在寻找的。​​p>