我有一个应用程序,我想要进行身份验证,但不使用用户模型。这个模型叫做Karyawan。但是,当我更改auth.php时,它开始显示以下错误:
Illuminate\Auth\EloquentUserProvider::validateCredentials():参数#1 ($user)必须类型为照明\Contracts\Auth\Authenticatable,App\Models\Karyawan给定,在第426行的C:\KKP\Project_KKP\vendor\laravel\framework\src\Illuminate\Auth\SessionGuard.php中调用
Laravel版本是8
auth.php
'defaults' => [
'guard' => 'web',
'passwords' => 'users',
],
'guards' => [
'web' => [
'driver' => 'session',
'provider' => 'users',
],
],
'providers' => [
'users' => [
'driver' => 'eloquent',
'model' => App\Models\Karyawan::class,
],
],
'passwords' => [
'users' => [
'provider' => 'users',
'table' => 'password_resets',
'expire' => 60,
'throttle' => 60,
],
],App\Models\Karyawan.php
namespace App\Models;
use App\Models\Cuti;
use App\Models\Role;
use App\Models\Divisi;
use App\Models\Jabatan;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Factories\HasFactory;
class Karyawan extends Model
{
use HasFactory;
protected $guarded = ['karyawan_id'];
protected $table = 'karyawans';
public function scopeSearch($query, array $searchs) {
$query->when($searchs['search'] ?? false, function($query, $search) {
return $query->where('nama', 'like', '%' . $search . '%')
->orWhere('divisi', 'like', '%' . $search . '%')
->orWhere('jabatan', 'like', '%' . $search . '%')
->orWhere('agama', 'like', '%' . $search . '%')
->orWhere('nik', 'like', '%' . $search . '%');
});
}
public function role_id()
{
return $this->belongsTo(Role::class);
}
public function divisi()
{
return $this->belongsTo(Divisi::class);
}
public function jabatan()
{
return $this->belongsTo(Jabatan::class);
}
public function cuti()
{
return $this->hasMany(Cuti::class);
}
}web.php
Route::get('/', [BasicLoginController::class, 'login']);
Route::post('/', [BasicLoginController::class, 'authenticate']);BasicLoginController.php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use Illuminate\Support\Facades\Auth;
class BasicLoginController extends Controller
{
public function login()
{
return view ('login.login', [
'title' => 'Login'
]);
}
public function authenticate(Request $request) {
$credentials = $request->validate([
'email' => 'required|email:dns',
'password' => 'required'
]);
if(Auth::attempt($credentials)) {
$request->session()->regenerate();
return redirect()->intended('/dashboard');
}
return back()->with('loginError', 'Email atau Password salah');
}
}发布于 2022-05-16 15:33:34
正如错误说明的那样,类需要实现接口Illuminate\Contracts\Auth\Authenticatable,公共用户模型通过扩展类Illuminate\Foundation\Auth\User as Authenticatable;来实现该接口。
因此,要么在类中实现契约,然后在类中声明所有需要的方法功能。
或
只需扩展现有的Illuminate\Foundation\Auth\User
namespace App\Models;
//...
use Illuminate\Foundation\Auth\User as Authenticatable;
class Karyawan extends Authenticatable
{类Illuminate\Foundation\Auth\User扩展了模型类Illuminate\Database\Eloquent\Model,因此模型功能仍然可用。
编辑:
在这里,复制粘贴这个模型,您应该学习一些PSRs,特别是PSR-4
namespace App\Models;
use App\Models\Cuti;
use App\Models\Role;
use App\Models\Divisi;
use App\Models\Jabatan;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Foundation\Auth\User as Authenticatable;
class Karyawan extends Authenticatable
{
use HasFactory;
protected $guarded = ['karyawan_id'];
protected $table = 'karyawans';
public function scopeSearch($query, array $searchs) {
$query->when($searchs['search'] ?? false, function($query, $search) {
return $query->where('nama', 'like', '%' . $search . '%')
->orWhere('divisi', 'like', '%' . $search . '%')
->orWhere('jabatan', 'like', '%' . $search . '%')
->orWhere('agama', 'like', '%' . $search . '%')
->orWhere('nik', 'like', '%' . $search . '%');
});
}
public function role_id()
{
return $this->belongsTo(Role::class);
}
public function divisi()
{
return $this->belongsTo(Divisi::class);
}
public function jabatan()
{
return $this->belongsTo(Jabatan::class);
}
public function cuti()
{
return $this->hasMany(Cuti::class);
}
}发布于 2022-05-16 15:33:47
类Karyawan需要实现Illuminate\Contracts\Auth\Authenticatableclass接口。或者可以在Karyawan类中扩展照明\Foundation\Auth\User。
https://stackoverflow.com/questions/72261708
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