Laravel 5: Check if a category collection has a valid result or no records at all

A very casual question with the multi level DB tables and eloquent query system.

A MainCategory has more than one Category(categories table)

A Category has more than one Product(products table)

Now the routes.php has:

Route::get('/categories/{slug}', ['as' => 'categories.list', 'uses' => 'ListController@getCategories'])->where('slug', '[\w\d\-\_]+');

The ListController@getCategories method wants to list all products along with the categories come under the main_category.

class ListController extends Controller
{
    public function getCategories($slug) {
        $categories = Category::orderBy('id', 'asc')
            ->select('categories.id as cat_id', 'categories.name as cat_name','categories.slug as cat_slug','products.*')
            ->join('main_categories', 'main_categories.id', '=', 'categories.main_category_id')
            ->join('products', 'products.category_id', '=', 'categories.id')
            ->where('main_categories.slug', '=', $slug)
            ->get();
        if(count($categories) >  0) {
            return view('pages.categories')->withCategories($categories);
        }
        else { // either main_categories.name not in $slug or there is no category available for this main_category

        }
    }
}

The code works fine as expected, listing both products and categories from the same collection object.

Now there is an else case. How could I check the result collection so that I can differentiate if there was a main_category in the name of given $slug or if there was no categories under the main_category with $slug.

See the code, count($categories) returns 0 for both cases. I don't stick on my eloquent way. You may suggest me another query or a condition check with PHP. I would like to know the best efficient way to do this.

[PS:All relations are direct and pretty basic(belongsTo and hasMany) and table names are in common standard- the Laravel 5 way of dooing it(using v5.2) so I guess you get if there is a code needed/missing in my question.]

via Chebli Mohamed