Php 在数据库mysql中创建子菜单时出错

我是拉雷维尔的新手。我想创建我的网站使用雄辩的子菜单。然而，它给了我错误的说法

找不到基表或视图：1146表“shopping.sub_menu”不存在（SQL:select*from

sub_menu

其中

sub_menu

菜单id=1且

sub_menu

菜单id不为空）

我的模型菜单 Menu.php

class Menu extends Model
{


   public function submenu()
   {
      return $this->hasMany(SubMenu::class);
   }
}

 // SubMenu Model
class SubMenu extends Model
{
   public function menu()
   {
      return $this->belongsTo(Menu::class);
   }
}

 <ul>
    @foreach ($menu as $_menu)
    <li>
        <a href="{{ route($_menu->route) }}" id={{ $_menu->id }} class="side-menu">
            @if ($_menu->submene->count())
            <ul>
                @foreach ($_menu->submenu as $_submenu)
                <li><a href="{{ route($_menu->route) }}" id={{ $_menu->id }}>{{$_submenu->$_menu}} </a></li>@endforeach
            <div class="side-menu__icon"> <i data-feather={{ $_menu->icon_name }}></i> </div>
            <div class="side-menu__title"> {{ $_menu->menu_name }} </div>
        </a>
    </li>
    @endforeach

下面是刀片 side nav.blade.php

class Menu extends Model
{


   public function submenu()
   {
      return $this->hasMany(SubMenu::class);
   }
}

 // SubMenu Model
class SubMenu extends Model
{
   public function menu()
   {
      return $this->belongsTo(Menu::class);
   }
}

 <ul>
    @foreach ($menu as $_menu)
    <li>
        <a href="{{ route($_menu->route) }}" id={{ $_menu->id }} class="side-menu">
            @if ($_menu->submene->count())
            <ul>
                @foreach ($_menu->submenu as $_submenu)
                <li><a href="{{ route($_menu->route) }}" id={{ $_menu->id }}>{{$_submenu->$_menu}} </a></li>@endforeach
            <div class="side-menu__icon"> <i data-feather={{ $_menu->icon_name }}></i> </div>
            <div class="side-menu__title"> {{ $_menu->menu_name }} </div>
        </a>
    </li>
    @endforeach

@foreach（$菜单作为$菜单）

@endforeach
图标\u名称}}>
{{$\菜单->菜单名}

@endforeach

愚蠢的问题：你真的创建了那个表吗？您是否运行了迁移？是的，我进行了迁移我相信您可以在模型中定义表名来解决此类问题

protected$table='sub_menu'