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laravel中的Auth :: login（$ user）无法登录用户

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我正在使用 laravel 5.4 和 Auth::login($user) 显示类型错误：

传递给Illuminate \ Auth \ SessionGuard :: login（）的参数1必须实现接口Illuminate \ Contracts \ Auth \ Authenticatable，给出的Illuminate \ Database \ Eloquent \ Builder实例，在/ home / vendor / laravel / framework / src /中调用第294行的Illuminate / Auth / AuthManager.php

我的User.php文件是：

namespace App;

use Illuminate\Notifications\Notifiable;
use Illuminate\Foundation\Auth\User as Authenticatable;
class User extends Authenticatable
{
use Notifiable;
/**
 * The attributes that are mass assignable.
 *
 * @var array
 */
protected $fillable = [
    'name', 'email', 'role', 'password',
];

/**
 * The attributes that should be hidden for arrays.
 *
 * @var array
 */
protected $hidden = [
    'password', 'remember_token',
];
}

我的AuthController.php是：

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use App\User;
use Auth;
use Illuminate\Support\Facades\Input;
use \Hash;

class AuthController extends Controller
{
//
public function register(Request $request){
    $this->validate($request, [
        'name' => 'required|max:30',
        'email' => 'required|email|unique:users',
        'regpassword' => 'required|alpha_num|confirmed'
        ]);

    /*$user = new User();
    $user['name'] = $request['name'];
    $user['email'] = $request['email'];
    $user['password'] = bcrypt($request['password']);
    $user['role'] = 'user';
    $user->save();
    */
    $user = User::create(array(
        'email' => Input::get('email'),
        'name' => Input::get('name'),
        'password' => Hash::make(Input::get('password')),
        'role' => 'user'));
    return 'success';


 }

public function userLogin(Request $request){
    $this->validate($request,[
        'email' => 'required|email',
        'password' => 'required' 
        ]);

    $user = User::where('email', '=', $request['email'])-> where('password' ,'=', Hash::make($request['password']))->where('role','=','user');                                                    

    if($user){
        Auth::login($user); 
        return $next($request);
    }
    else{
        return redirect()->back();
    }


}
}

我的web.php

Route::get('/', function () {
return view('welcome');
});

Auth::routes();

Route::get('/home', 'HomeController@index');
Route::post('/register', 'AuthController@register');
Route::post('/UserLogin','AuthController@userLogin');
laravel authentication laravel-5.2 laravel-5.3 laravel-5.4

3 回答

  • 1

    你的问题是这样的：

    $user = User::where('email', '=', $request['email'])-> where('password' ,'=', Hash::make($request['password']))->where('role','=','user');

    缺少从查询中获取结果集的调用 . 添加 ->first() .

    它的意思是：

    您给了我一个查询构建器的实例

    但

    我想要一个扩展可验证类的模型

    如果你打开 App\User ，你会发现它确实扩展了这个类：

    class User extends Authenticatable {
    回复于
  • 0

    这个：

    $user = User::where('email', '=', $request['email'])-> where('password' ,'=', Hash::make($request['password']))->where('role','=','user');

    不是你获得用户的方式 .

    代替

    $user = User::where('email', '=', $request['email'])-> where('password' ,'=', Hash::make($request['password']))->where('role','=','user');                                                    

if($user){
    Auth::login($user); 
    return $next($request);
}
else{
    return redirect()->back();
}

    你可以使用：

    $logged = auth()->attempt(
          ['email' => $request['email'], 
           'password' => $request['password'], 
           'role' => 'user']);

    if($logged) {
        return $next($request); // probably it won't work. This is fine in middleware but not in controller
    }
    else{
        return redirect()->back();
    }
    回复于
  • 0

    解决了我的问题如下：

    $user = User::where('email', '=' $request['email'])->where('role','=','user')->first();
if(Hash::check($request['password'],$user->password)){
    Auth::login($user);
    return  'success';
}
    回复于

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