如何在模板视图中的laravel上创建子菜单？

Question

i want to create a sub menu on my templates/master. i want all my process in there. i have table: menu , role , and menu_role (conatins many to many from menu and role).

i tried to select table menu through menu_role, because i want display menu according to its role. it success but there is no sub menu. i'm bad at query builder with relationship.

this is my templates/master.blade.php :

use App\MenuRole;
use App\Role;
use App\Menu;
use App\Admin;
use App\Operator;

$userCheck = Auth::user()->role_id;
$menus = MenuRole::with('menu')->where('role_id', $userCheck)->get();

and for the html, if there is no module on table menu i want to display this :

@foreach($menus as $menu)
    <li><a href="/{{ Auth::user()->role->role_name }}/{{ $menu->menu->route }}" target="{{ $menu->menu->target }}">{{ $menu->menu->title }}</a></li>
@endforeach

and if there is a module , i want to display this :

<li class="has-sub">
      <a class="js-arrow" href="#">
          <i class="fas fa-tachometer-alt"></i>{{ $menu->module }}</a>
      <ul class="navbar-mobile-sub__list list-unstyled js-sub-list">
          <li><a href="/{{ Auth::user()->role->role_name }}/{{ $menu->menu->route }}">{{ $menu->menu->title }}</a></li>
      </ul>
</li>

so the thing is, in table menu i have field module . this field will distinguish which ones have subMenu and which have no subMenu. and i want to foreach table menu. i have configure the model.

Answer 1

Will a simple @empty condition suffice? When a Menu has the module field filled it shows one thing, otherwise the other.

@foreach ($menus as $menu)
    @empty($menu->menu->module)

    @else

    @endempty
@endforeach

If this does not solve your problem, could you please go into more detail about what you want it to look like?