未找到角色错误

我正在尝试使用BjyAuthorize安装ZfcUser。尝试访问守护路线“/ user”时出现以下错误。未找到角色错误

Fatal error: Uncaught exception 'Zend\Permissions\Acl\Exception\InvalidArgumentException' with message 'Role '3' not found' in /home/brian/dev/ptapp/app/vendor/zendframework/zendframework/library/Zend/Permissions/Acl/Role/Registry.php:106 

Stack trace: 

#0 /home/brian/dev/ptapp/app/vendor/zendframework/zendframework/library/Zend/Permissions/Acl/Role/Registry.php(67): Zend\Permissions\Acl\Role\Registry->get('3') 

#1 /home/brian/dev/ptapp/app/vendor/zendframework/zendframework/library/Zend/Permissions/Acl/Acl.php(112): Zend\Permissions\Acl\Role\Registry->add(Object(Zend\Permissions\Acl\Role\GenericRole), Array) 

#2 /home/brian/dev/ptapp/app/vendor/bjyoungblood/bjy-authorize/src/BjyAuthorize/Service/Authorize.php(277): Zend\Permissions\Acl\Acl->addRole('bjyauthorize-id...', Array) 

#3 /home/brian/dev/ptapp/app/vendor/bjyoungblood/bjy-authorize/src/BjyAuthorize/Service/Authorize.php(90): BjyAuthorize\Service\Authorize->load() 

#4 /home/brian/dev/ptapp/app/vendor/bjyoungblood/bjy-authorize/src/BjyAuthorize/Service/Authorize.php(239) in /home/brian/dev/ptapp/app/vendor/zendframework/zendframework/library/Zend/Permissions/Acl/Role/Registry.php on line 69

我的数据库表被用在bjyauthorize库所提供的schema.sql文件，其中包含下面的代码创建：

CREATE TABLE IF NOT EXISTS `user_role` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `roleId` varchar(255) NOT NULL, 
    `is_default` tinyint(1) NOT NULL, 
    `parent_id` varchar(255) DEFAULT NULL, 
    PRIMARY KEY (`id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

CREATE TABLE IF NOT EXISTS `user_role_linker` (
    `user_id` int(11) unsigned NOT NULL, 
    `role_id` int(11) NOT NULL, 
    PRIMARY KEY (`user_id`,`role_id`), 
    KEY `role_id` (`role_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 

;

的条目：

user_role

id | roleId | is_default | parent_id 
--------------------------------------------- 
1  admin   0   therapist 
2  therapist  0   patient 
3  patient   0   user 
4  guest   1   NULL 
5  user   0   NULL

user_role_linker

user_id  role_id 
------------------- 
    3   3

我不得不修改我的bjyauthorize.global.php文件，因为“role_id_field”和“parent_role_field”值与提供的sql schema不匹配。它看起来像这样：

<?php 

return array(
    'bjyauthorize' => array(

     'default_role' => 'guest', 

     'identity_provider' => 'BjyAuthorize\Provider\Identity\ZfcUserZendDb', 

     'role_providers' => array(
      'BjyAuthorize\Provider\Role\ZendDb' => array(
       'table'    => 'user_role', 
       'role_id_field'  => 'roleId', 
       'parent_role_field' => 'parent_id', 
      ), 
     ), 

     'guards' => array(
      'BjyAuthorize\Guard\Route' => array(
       array('route' => 'zfcuser', 'roles' => array('user')), 
       array('route' => 'zfcuser/logout', 'roles' => array('user')), 
       array('route' => 'zfcuser/login', 'roles' => array('guest')), 
       array('route' => 'zfcuser/register', 'roles' => array('guest')), 
       // Below is the default index action used by the ZendSkeletonApplication 
       array('route' => 'home', 'roles' => array('guest', 'user')), 
      ), 
     ), 
    ), 
);

而且我application.config.php看起来是这样的：

<?php 
return array(
    'modules' => array(
     'Application', 
     'ZfcBase', 
     'ZfcUser', 
     'User', 
     'BjyAuthorize', 
    ), 

    'module_listener_options' => array(
     'module_paths' => array(
      './module', 
      './vendor', 
     ), 

     'config_glob_paths' => array(
      'config/autoload/{,*.}{global,local}.php', 
     ), 
    ), 
);

凡模块 “用户” 我的自定义ZfcUser实体。

问题出在哪里？我对ZF2和zfc很新，所以感谢您的帮助！

编辑：我应该注意到，当我没有登录，我正确地得到403时，试图访问/用户，这是当我登录，我收到此消息。我试图访问任何路由时收到它（除无效路由，我得到一个404）

来源

2013-08-17 Brian

该问题似乎与提供的sql模式。在user_role_linker表中，将role_id字段更改为VARCHAR。在输入这个表时，user_id应该对应于用户的数字ID，但是role_id应该是user_role表中“roleId”字段的文本ID。

来源

2013-08-17 15:26:05 Brian

基于第一注释，如下的sql作品，在PARENT_ID存在同样的问题，你需要改变它，太：

CREATE TABLE IF NOT EXISTS `user_role` (
    `id` INT(11) NOT NULL AUTO_INCREMENT, 
    `role_id` VARCHAR(255) NOT NULL, 
    `is_default` TINYINT(1) NOT NULL DEFAULT 0, 
    `parent_id` VARCHAR(255) NULL, 
    PRIMARY KEY (`id`), 
    UNIQUE INDEX `unique_role` (`role_id` ASC), 
    INDEX `idx_parent_id` (`parent_id` ASC), 
    CONSTRAINT `fk_parent_id` FOREIGN KEY (`parent_id`) REFERENCES `user_role` (`role_id`) ON DELETE SET NULL 
) ENGINE = InnoDB DEFAULT CHARACTER SET = utf8 COLLATE = utf8_unicode_ci; 

CREATE TABLE IF NOT EXISTS `user_role_linker` (
    `user_id` INT UNSIGNED NOT NULL, 
    `role_id` VARCHAR(255) NOT NULL, 
    PRIMARY KEY (`user_id`, `role_id`), 
    INDEX `idx_role_id` (`role_id` ASC), 
    INDEX `idx_user_id` (`user_id` ASC), 
    CONSTRAINT `fk_role_id` FOREIGN KEY (`role_id`) REFERENCES `user_role` (`role_id`) ON DELETE CASCADE, 
    CONSTRAINT `fk_user_id` FOREIGN KEY (`user_id`) REFERENCES `user` (`user_id`) ON DELETE CASCADE 
) ENGINE = InnoDB DEFAULT CHARACTER SET = utf8 COLLATE = utf8_unicode_ci;

来源

2013-10-06 13:30:40 user2851823

未找到角色错误

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