laravel 5.5中的嵌套查询

Question

我想从products

获取AppServiceProvider信息

逻辑

类别 - ＆gt;子类别 - ＆gt;产品

现在我需要products基于类别id

注意：您可能认为这很奇怪，但事实并非如此，只是想象您希望在该类别的所有子类别的类别页面中展示产品。那你就像我一样需要这样的事情。在category_id中获取产品时，只在产品表中保存subcategory_id。

这是AppServiceProdiver中的循环：

View::composer('welcome', function ($view) {
          $categories = Category::join('admin_designs', 'admin_designs.category_id', '=', 'categories.id')->get();
          foreach($categories as $category){
            $subcategory = Subcategory::join('categories', 'categories.id', '=', 'subcategories.category_id')->first();
            $designs = Product::where('subcategory_id', $subcategory)->first();
          }
        $view->with('categories', $categories);
        });

该循环的结果是：

{"id":2,"title":null,"slug":"laptop","image":"category-1516430091.jpg","imageAlt":null,"status_id":1,"meta_tags":"laptop,tags","meta_description":"laptop description","created_at":"2018-02-01 11:41:30","updated_at":"2018-02-01 11:41:30","category_id":1},

  PS：不知道那是什么！不是产品/不完整   类别less title! ...：|

任何人都可以帮助修复该查询吗？

Answer 1

解决

View::composer('welcome', function ($view) {
            $category = DB::table('admin_designs')
                              ->join('categories', 'categories.id', '=', 'admin_designs.category_id')
                              ->join('subcategories', 'subcategories.category_id', '=', 'categories.id')
                              ->join('products', 'products.subcategory_id', '=', 'subcategories.id')
                                ->get();

            $view->with('category', $category);
        });

希望它有所帮助。