未定义$ user-> withAccessToken函数/方法

Question

我在laravel中编写了一个API代码，在这里我的代码是在我使用护照的这个API中

class TransactionApi extends BaseController
{
    public function __construct()
    {
        $this->middleware('client');
    }

在构造函数中，我称为客户端中间件

kernel.php

protected $routeMiddleware = [
    ...
    'client' => CheckClientCredentials::class,
];

这是我的API服务提供商

protected function mapApiRoutes()
{
    Route::prefix('api/v1')
        ->middleware('api')
        ->namespace($this->namespace)
        ->group(__DIR__ . '/../routes/api.php');
}

API路由

Route::group(['prefix' => 'transaction'], function () {
    Route::get('all', 'TransactionApi@showAllWithPaginate');
    Route::get('total_refund', 'TransactionApi@totalRefund');
    Route::get('total_price', 'TransactionApi@totalPrice');
    Route::get('/', 'TransactionApi@show');
});

用户模型

<?php

namespace Corals\User\Models;

use Corals\Foundation\Traits\Hookable;
use Corals\Foundation\Traits\HashTrait;
use Corals\Foundation\Transformers\PresentableTrait;
use Corals\Modules\CMS\Models\Content;
use Corals\Settings\Traits\CustomFieldsModelTrait;
use Illuminate\Notifications\Notifiable;
use Illuminate\Foundation\Auth\User as Authenticatable;
use Spatie\Activitylog\Traits\LogsActivity;
use Spatie\MediaLibrary\HasMedia\HasMediaTrait;
use Spatie\MediaLibrary\HasMedia\Interfaces\HasMediaConversions;
use Spatie\MediaLibrary\Media;
use Spatie\Permission\Traits\HasRoles;
use Yajra\Auditable\AuditableTrait;
use Corals\User\Traits\TwoFactorAuthenticatable;
use Corals\User\Contracts\TwoFactorAuthenticatableContract;
use Corals\User\Notifications\MailResetPasswordNotification;
use Corals\Modules\Subscriptions\Models\Subscription;
use Laravel\Passport\HasApiTokens;

class User extends Authenticatable implements HasMediaConversions, TwoFactorAuthenticatableContract
{
    use
        TwoFactorAuthenticatable,
        HasApiTokens,
        Notifiable,
        HashTrait,
        HasRoles,
        Hookable,
        PresentableTrait,
        LogsActivity,
        HasMediaTrait,
        AuditableTrait,
        CustomFieldsModelTrait;

    protected $slack_url = null;

    /**
     *  Model configuration.
     * @var string
     */
    public $config = 'user.models.user';

    protected static $logAttributes = ['name', 'email'];
    protected static $ignoreChangedAttributes = ['remember_token'];

    protected $casts = [
        'address' => 'json',
        'notification_preferences' => 'array'
    ];

    protected $appends = ['picture', 'picture_thumb'];

    protected $dates = ['trial_ends_at'];

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $guarded = [
        'id'
    ];

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

    public function __construct(array $attributes = [])
    {
        $config = config($this->config);

        if (isset($config['presenter'])) {
            $this->setPresenter(new $config['presenter']);
            unset($config['presenter']);
        }

        foreach ($config as $key => $val) {
            if (property_exists(get_called_class(), $key)) {
                $this->$key = $val;
            }
        }

        return parent::__construct($attributes);
    }

鉴于此设置，可以在本地进行测试，它将返回请求的json

但他们在登台环境中的设置相同，返回了

{
  "message": "Call to undefined method Illuminate\\Auth\\GenericUser::withAccessToken()",
  "exception": "Symfony\\Component\\Debug\\Exception\\FatalThrowableError",
  "file": "/path/to/vendor/laravel/passport/src/Guards/TokenGuard.php",
  "line": 148,

登台的可能原因是什么

在此行

return $token ? $user->withAccessToken($token) : null;

未定义的$ user-> withAccessToken函数/方法？