php Laravel-5 向 Auth 添加 hasRole 方法
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Laravel-5 adding hasRole method to Auth
提问by V4n1ll4
I'm trying to extend the Laravel-5.1 Authmiddleware so that I can add my own method to it:
我正在尝试扩展 Laravel-5.1Auth中间件，以便我可以向其中添加我自己的方法：
Auth::hasRole()
Auth::hasRole()
What do I need to do in order to add the new method hasRole to Auth?
我需要做什么才能将新方法 hasRole 添加到 Auth？
Here is my routes file:
这是我的路由文件：
/* Administrator Routes */
Route::group(['namespace' => 'Admin', 'middleware' => 'timesheets.admin:Administrator'], function()
{
Route::get('home', 'AdminController@index');
});
Here is my middleware file:
这是我的中间件文件：
<?php
namespace App\Http\Middleware\Auth;
use Closure;
use Illuminate\Contracts\Auth\Guard;
class AdminAuthenticate
{
/**
* The Guard implementation.
*
* @var Guard
*/
protected $auth;
/**
* Create a new filter instance.
*
* @param Guard $auth
* @return void
*/
public function __construct(Guard $auth)
{
$this->auth = $auth;
}
/**
* Handle an incoming request.
*
* @param \Illuminate\Http\Request $request
* @param \Closure $next
* @return mixed
*/
public function handle($request, Closure $next, $role)
{
if ($this->auth->guest()) {
if ($request->ajax()) {
return response('Unauthorized.', 401);
} else {
return redirect()->guest('login');
}
}
if (auth()->check() && auth()->user()->hasRole($role)) {
return $next($request);
}
}
}
回答by jakehallas
Could you try adding the following to your User model:-
您能否尝试将以下内容添加到您的用户模型中：-
public function hasRole($role)
{
return User::where('role', $role)->get();
}
This should firstly check to see if you User table has the field 'role' and then check your parameter $roleagainst the rolefield.
这应该首先检查您的用户表是否具有字段“角色”，然后$role根据该role字段检查您的参数。
You can the check by doing the following:
您可以通过执行以下操作进行检查：
if( Auth::user()->hasRole($role) )
You may need to adjust the example to your needs. Let me know if you need anything else.
您可能需要根据需要调整示例。需要帮助请叫我。
/------------EDIT-----------------/
/ - - - - - - 编辑 - - - - - - - - -/
If you have two seperate tables, one holding the user information and the other holding the users privileges/roles you could add another function to the User model:
如果您有两个单独的表，一个保存用户信息，另一个保存用户权限/角色，您可以向 User 模型添加另一个功能：
public function userID()
{
return $this->user_id;
}
This will check for if you have a user ID field if so, it will return the id for the authenticated user.
这将检查您是否有用户 ID 字段，如果有，它将返回经过身份验证的用户的 ID。
Then add this to your hasRolesmethod:
然后将其添加到您的hasRoles方法中：
public function hasRoles($userID, $roles)
{
return Your\User\Roles\Model::where('role', $role)->where('user_id', $user_id)->get();
}
Your middleware would look like this:
你的中间件看起来像这样：
public function handle($request, Closure $next, $role)
{
if ($this->auth->guest()) {
if ($request->ajax()) {
return response('Unauthorized.', 401);
} else {
return redirect()->guest('login');
}
}
$userID = Auth::user()->userID();
if (auth()->check() && auth()->user()->hasRole($userID, $role)) {
return $next($request);
}
}
If I understood correctly what you want. I believe this should work for you.
如果我理解正确你想要什么。我相信这应该对你有用。
回答by mike
I've taken a different tack by using a trait in my Usermodel.
我通过在我的User模型中使用特征采取了不同的策略。
<?php
namespace App\Traits;
use App\Role;
use App\User;
trait HasRoles{
public function roles()
{
return $this->belongsToMany('App\Role');
}
public static function findByRole(Role $role)
{
return $role->users()->get();
}
public function hasRole(Role $role)
{
return $this->roles()->get()->contains($role);
}
}
回答by omarjebari
There are some good packages to help with this if you don't want to brew your own. I can recommend both: Zizaco Entrust: https://github.com/Zizaco/entrustand Sentinel: https://cartalyst.com/manual/sentinel/2.0
如果你不想自己酿造，有一些很好的软件包可以帮助解决这个问题。我可以同时推荐：Zizaco Entrust：https: //github.com/Zizaco/entrust和 Sentinel：https://cartalyst.com/manual/sentinel/2.0
