如何在laravel中不使用foreach循環顯示動態值

Question

如何在laravel中打印這些類型的數據。我試圖創建動態菜單併發生這些類型的問題。代碼就是這樣。我的代碼是這樣的。 鑑於作曲家。

public function compose(View $view) 
{ 
    $menus = $this->menu->getDynamicMenus(); 

    $user_role = $this->role->lists('name', 'id'); 
    //dd($user_role); 
    $view->with('user_role', $user_role)->with('menus', $menus); 
} 

menurepository code. 

public function getDynamicMenus() 
{ 
    $user = $this->auth->user(); 

    $roles = $user->roles; 

//  dd($roles); 
     $menus = []; 
     foreach ($roles as $role) { 
//dd($role); 
      foreach ($role->menus as $menu) { 

       if ($menu["parent_id"] > 0) { 
        $menus[$menu["parent_id"]][$menu["id"]] = $menu->toArray(); 
       } else { 
        $menus[$menu["id"]] = $menu->toArray(); 
       } 
      } 
     } 
     // dd($menus); 
     return $menus; 
    } 


and i print in frontend like. 

     @foreach($menus as $menu) 

        {{-- {!! Form::open() !!} 
        {!! Form::select('menus',$menus) !!} 
             {!! Form::close() !!}--}} 
        {{--{!! $menuitem->menu_name !!}--}} 
        {!! $menu->menu_url !!} 


        @endforeach 

，它顯示了問題：

Trying to get property of non-object (View: E:\xampp\htdocs\basicwc\resources\views\layouts\partials\sidebar.blade.php) (View: E:\xampp\htdocs\basicwc\resources\views\layouts\partials\sidebar.blade.php) (View: E:\xampp\htdocs\basicwc\resources\views\layouts\partials\sidebar.blade.php) 

as menu are getting in array like in image. 
any help?? 

我試圖做這樣的：

{!! menu($menus) !!} 

but it says undefined problem. 

我增加了一個菜單，它的結構是這樣的：

Answer 1

訪問它像一個數組：

@foreach($menus as $menu) 

... 
{{ $menu['menu_name'] }} 
{{ $menu['menu_url'] }} 

@endforeach