doq70020 2014-12-23 03:31
浏览 53
已采纳

Laravel Blade无法使用Sentry方法

I'm trying to write a .blade.php view that does:

  • Show the user profile information
  • If the user is banned, show a message "User is banned." instead.
  • If you're logged in as an ADMIN (that is, if you have the permission 'admin'), instead of showing that message, show the user profile with an extra information saying the user is banned.

    Here is my view:

    @extends ('layout.main')
    
    @section('content')
    
    @if(! $throttle->banned==1)
            {{ $user->username }} <br>
            {{ $user->summoner_name }}<br><br>
            {{ $user->bio }}<br>
            {{ $user->punishments }}<br>
    @else
            {{ "This user is banned." }}
                    @if(Sentry::getUser()->hasAccess('admin'))
                            {{ $user->username }} <br>
                            {{ $user->summoner_name }}<br><br>
                            {{ $user->bio }}<br>
                            {{ $user->punishments }}<br>
                            {{ ($throttle->banned == 1) ? "User is banned." : '' }}<br>
                            {{ $throttle->getSuspensionTime() }}<br>
                     @endif
    @endif
    
    @stop
    

And my controller:

<?php

class ProfileController extends BaseController {

        public function main($username) {

                try {
                        $throttle = Sentry::findThrottlerByUserLogin($username);

                        $user = User::where('username','=',$username);

                        if($user->count()) {

                                $user = $user->first();
                                return View::make('profile.main')
                                        ->with('user',$user)
                                        ->with('throttle',$throttle);

                        } else {

                                return App::abort(404);
                        }
                }
                 catch (Cartalyst\Sentry\Users\UserNotFoundException $e)
                 {
                        echo 'User not found.';
                 }
        }

}

That said, I keep getting the error "Call to a member function hasAccess() on a non-object" on the debugger whenever I use

@if(Sentry::getUser()->hasAccess('admin'))

The syntax looks OK to me, so I really don't get why Sentry does not recognize it as an object. What could it be? Is there a simple solution?

Thanks in advance!

  • 写回答

2条回答 默认 最新

  • duanmianhong4893 2014-12-23 16:50
    关注

    Ok, problem solved.

    @patricus, thanks for the help! That was not what I meant but it gave me the info I needed. When I called the method Sentry::getUser(), I was trying to get the user who was currently logged in and seeing the view, not the profile owner. But as you said, the method was returning NULL because I used it without checking if there was a logged in user to begin with!

    So what actually fixed the issue was rewriting

    @if(Sentry::getUser()->hasAccess('admin'))

    to

    @if(Sentry::check() && Sentry::getUser()->hasAccess('admin'))

    And thank you for the extra tips. I'm fairly new to Sentry, Laravel and PHP itself, so the code may get messy. All fixed and cleaned.

    Thanks!

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥15 luckysheet
  • ¥15 ZABBIX6.0L连接数据库报错,如何解决?(操作系统-centos)
  • ¥15 找一位技术过硬的游戏pj程序员
  • ¥15 matlab生成电测深三层曲线模型代码
  • ¥50 随机森林与房贷信用风险模型
  • ¥50 buildozer打包kivy app失败
  • ¥30 在vs2022里运行python代码
  • ¥15 不同尺寸货物如何寻找合适的包装箱型谱
  • ¥15 求解 yolo算法问题
  • ¥15 虚拟机打包apk出现错误