php Laravel 5.2：使用`Zizaco/entrust` 时如何获取当前用户的角色？
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Laravel 5.2 : How to get the role of current user when using `Zizaco/entrust`?
提问by zwl1619
I am using Laravel 5.2 and Zizaco/entrust5.2,my question is:
How to get the role of current user when using Zizaco/entrust?
我正在使用 Laravel 5.2 和Zizaco/entrust5.2，我的问题是：
如何在使用时获取当前用户的角色Zizaco/entrust？
NameAndRole.php
NameAndRole.php
namespace App\Services;
use App\User;
use App\Role;
use Zizaco\Entrust\EntrustRole;
use Illuminate\Support\Facades\Cache;
class NameAndRole
{
public $username;
public $role;
public function __construct() {
$user = \Auth::user();
$this->username = $user->name;
$this->role =$user->roles->first()->name;
}
}
Models:
楷模：
Role:https://github.com/Zizaco/entrust#role
角色：https: //github.com/Zizaco/entrust#role
<?php namespace App;
use Zizaco\Entrust\EntrustRole;
class Role extends EntrustRole
{
}
User:https://github.com/Zizaco/entrust#user
用户：https: //github.com/Zizaco/entrust#user
<?php
use Zizaco\Entrust\Traits\EntrustUserTrait;
class User extends Eloquent
{
use EntrustUserTrait; // add this trait to your user model
...
}
view:sidebar.blade.php
视图：sidebar.blade.php
@inject('details','App\Services\NameAndRole')
{{ $details->username }}
{{ $details->role }}
error:
错误：
ErrorException in NameAndRole.php line 20:
Trying to get property of non-object (View: D:\wnmp\www\laravel-entrust\resources\views\employer\partials\sidebar.blade.php) (View: D:\wnmp\www\laravel-entrust\resources\views\employer\partials\sidebar.blade.php)
回答by haakym
You can access it like any other relationship in eloquent. So to access the roles of a user you can do:
您可以像 eloquent 中的任何其他关系一样访问它。因此，要访问用户的角色，您可以执行以下操作：
$user->roles
This returns an eloquent collection as a Usercan have many Roles (i.e. it won't be just one), so if you're only expecting one role and you want the string value of it, you could do:
这将返回一个 eloquent 集合，因为 aUser可以有多个Roles（即它不会只有一个），所以如果您只期待一个角色并且您想要它的字符串值，您可以这样做：
$user->roles->first()->name // or display_name
I guess you could cast it to an array using toArray()method, if you wanted to work with an array instead:
我想你可以使用toArray()方法将它转换为数组，如果你想使用数组来代替：
$user->roles->toArray()
Edit
编辑
You could check the user has a role before assigning it:
您可以在分配之前检查用户是否具有角色：
$this->role = $user->roles ? $user->roles->first()->name : 'No role';
回答by Bharath N
To get the roles idfield:
要获取角色id字段：
$user->roles->first()->id;
