如何修复laravel 5.2 zizaco委托:迁移类名称验证？

Question

我从GitHub Link跟踪了zizac/entrust安装教程并遇到了错误:

类名必须是一个有效的对象或在VAR的字符串/ WWW/HTML/laravel_test /供应商/ zizaco /委托/ SRC /命令/ MigrationCommand.php上线86

MigrationCommand.php文件网址:链接

Outut:

php artisan entrust:migration

Tables: roles, role_user, permissions, permission_role
A migration that creates 'roles', 'role_user', 'permissions', 'permission_role' tables will be created in database/migrations directory

Proceed with the migration creation? [Yes|no] (yes/no) [yes]: yes

Creating migration...
PHP Fatal error:  Class name must be a valid object or a string in /var/www/html/laravel_test/vendor/zizaco/entrust/src/commands/MigrationCommand.php on line 86

命令:php artisan vendor:发布成功.

文件:config/entrust.php存在.

我没有改变config/auth.php文件的任何选项 - auth.php.怎么解决？

Answer 1

在第86行的vendor/zizaco/entrust/src/commands/MigrationCommand.php中

删除行:

    $usersTable  = Config::get('auth.table');
    $userModel   = Config::get('auth.model');

添加行:

$usersTable  = Config::get('auth.providers.users.table');
$userModel   = Config::get('auth.providers.users.model');

和config/auth.php文件一样写提供者行,就像我一样:

'providers' => [
    'users' => [
        'driver' => 'eloquent',
        'model' => App\User::class,
        'table' => 'users',
    ],

    // 'users' => [
    //     'driver' => 'database',
    //     'table' => 'users',
    // ],
],

然后你的问题将解决:快乐的编码

Answer 2

在第86行的vendor/zizaco/entrust/src/commands/MigrationCommand.php中.

Laravel 5.1.*添加行

$usersTable  = Config::get('auth.table');
$userModel   = Config::get('auth.model');

Laravel 5.2.*添加线

$usersTable  = Config::get('auth.providers.users.table');
$userModel   = Config::get('auth.providers.users.model');