如何解决尝试获取非对象的属性“sub_cat_name”

Question

通过使用视图作曲家传递变量，我得到了这样的错误：

“试图获取非对象的属性 'sub_cat_name'（查看：D:\xampp\htdocs\E-commerce\resources\views\layouts\includes\top.blade.php）”。请解决问题，代码如下：

AppServiceProvider：

View::composer(['layouts.includes.top'], function ($view) {
    $view->with('subcat', SubCategory::get());
});

控制器：

public function prdtview()
{

    $categories = Category::get();
    $data=DB::select('select category.cat_name,product_images.prdt_image,product.prdt_name,product.actual_price from product INNER JOIN category on product.catid = category.id INNER JOIN product_images on product.id = product_images.prdt_id where product.sub_cat_id = 3 ');

        return view("frontend.product",[
        'data' => $data,
        'category' =>$categories
    ]);
}

刀片文件：

 <div class="nav-item">
        <div class="container">
            <nav class="nav-menu mobile-menu">
                <ul>
                    <li class="active"><a href="">Home</a></li>
                    @foreach($category as $category)
                        <li><a href="#">{{$category->cat_name}}</a>
                        <ul class="dropdown">
                            @foreach($subcat as $subcat)
                            <li><a href="#"><?php echo $subcat->sub_cat_name; ?></a></li>
                            @endforeach
                        </ul>
                    </li>
                    @endforeach                        
                </ul>
            </nav>
            <div id="mobile-menu-wrap"></div>
        </div>
    </div>

Answer 1

改变它：

@foreach($subcat as $subcat)
      <li><a href="#"><?php echo $subcat->sub_cat_name; ?></a></li>
@endforeach

类似于：

 @foreach($subcat as $sub_cat)
     <li><a href="#"><?php echo $sub_cat->sub_cat_name; ?></a></li>
 @endforeach