我有一个使用<?php
namespace App\Http\Middleware;
use Closure;
use Menu;
use App\Module as modelmodule;
use auth;
class Frontend
{
public function handle($request, Closure $next) {
Menu::make('sidebar', function($menu){
$menu->add('Main Menu',array('class' => 'header'));
//USER//
$menu->add("User Control",array('nickname' => "user",'class'=>'treeview'))
->append(' <b class="caret"></b>')
->prepend('<span class="glyphicon glyphicon-user"></span> ');
$menu->user->add('Daftar User','user/list');
$menu->user->add('Tipe User','user/type');
//Product//
$menu->add("Product",array('nickname' => "product",'class'=>'treeview'))
->append(' <b class="caret"></b>')
->prepend('<span class="glyphicon glyphicon-list"></span> ');
$menu->product->add('Product List','user/list');
//About//
$menu->add("About Us",array('nickname' => "about"))
->prepend('<span class="glyphicon glyphicon-list"></span> ');
});
return $next($request);
// }
}
}
调用的中间件，该中间件的使用显示我使用laravel-menu创建的菜单
auth condition所以，我的问题是如何添加 if(auth()->user()->isDeveloper()) {
$menu->add("About Us",array('nickname' => "about"))
->prepend('<span class="glyphicon glyphicon-list"></span> ');
}
例如
auth但上面的class User extends Authenticatable
{
use Notifiable;
protected $fillable = [
'name', 'email', 'password','type','subscribestatus','usercode'
];
protected $hidden = [
'password', 'remember_token',
];
public function isDeveloper()
{
return ($this->type == 'TP001');
}
}
脚本无效。我收到此错误
在null
上调用成员函数isDeveloper（）
这是我的用户模型
number = benzin_txt.text as Number;
如何修复错误？对不起，我的英语不好 。
答案 0 :(得分：1)
确保用户在尝试调用其上的函数之前已登录：
if (auth()->check() && auth()->user()->isDeveloper()) {