如何为用户添加角色? Yii2

时间:2014-01-23 12:40:00

标签: php yii2

我们使用了Yii2框架的最后一个alpha。已创建用户的角色,但问题是它如何分配给用户。没有文件。

5 个答案:

答案 0 :(得分:15)

对于RBAC的数据库版本,请使用DbManager(quote frm:Alexufo):

use yii\rbac\DbManager;
$r=new DbManager;
$r->init();
$r->createRole("admin","Administrator");
$r->save();

$r->assign('1','admin');   //1 is user id 

示例访问规则:

<?php
namespace backend\controllers;

use yii;
use yii\web\AccessControl;
use yii\web\Controller;

class SiteController extends Controller
{
    public function behaviors()
    {
        return [
            'access' => [
                'class' => AccessControl::className(),
                'rules' => [
                    [
                        //'actions' => ['login', 'error'], // Define specific actions
                        'allow' => true, // Has access
                        'roles' => ['@'], // '@' All logged in users / or your access role e.g. 'admin', 'user'
                    ],
                    [
                        'allow' => false, // Do not have access
                        'roles'=>['?'], // Guests '?'
                    ],
                ],
            ],
        ];
    }

    public function actionIndex()
    {
        return $this->render( 'index' );
    }
}
?>

不要忘记将其添加到您的配置文件(config / main.php):

'components' => [
    'authManager'=>array(
        'class' => 'yii\rbac\DbManager',
        'defaultRoles' => ['end-user'],
    ),
    ...
]

表:

drop table if exists `tbl_auth_assignment`;
drop table if exists `tbl_auth_item_child`;
drop table if exists `tbl_auth_item`;

create table `tbl_auth_item`
(
   `name`                 varchar(64) not null,
   `type`                 integer not null,
   `description`          text,
   `biz_rule`              text,
   `data`                 text,
   primary key (`name`),
   key `type` (`type`)
) engine InnoDB;

create table `tbl_auth_item_child`
(
   `parent`               varchar(64) not null,
   `child`                varchar(64) not null,
   primary key (`parent`,`child`),
   foreign key (`parent`) references `tbl_auth_item` (`name`) on delete cascade on update cascade,
   foreign key (`child`) references `tbl_auth_item` (`name`) on delete cascade on update cascade
) engine InnoDB;

create table `tbl_auth_assignment`
(
   `item_name`            varchar(64) not null,
   `user_id`              varchar(64) not null,
   `biz_rule`              text,
   `data`                 text,
   primary key (`item_name`,`user_id`),
   foreign key (`item_name`) references `tbl_auth_item` (`name`) on delete cascade on update cascade
) engine InnoDB;

您还可以在“yii / rbac”目录(包括其他SQL文件)中找到此信息。 有关功能和更多详细信息:

https://github.com/yiisoft/yii2/blob/master/docs/guide/security-authorization.md

答案 1 :(得分:5)

$user_id = 1;

$auth = new DbManager;
$auth->init();
$role = $auth->createRole('editor');
$auth->add($role);

$auth->assign($role, $user_id);

=============================================== ========================== 如果你想选择角色而不是创建

$auth = new DbManager;
$auth->init();
$role = $auth->getRole('admin');
$auth->assign($role, $user_id);

100%工作!

答案 2 :(得分:3)

解决!

================创建角色============

use yii\rbac\PhpManager;
$r=new PhpManager;
$r->init();
$r->createRole("admin","Администратор"); 
$r->save();

=============== assign ==================

$r->assign('1','admin');   //1 is user id

答案 3 :(得分:1)

实现管理员角色的一种非常简单的方法是将其添加到您的控制器:

use yii;
/**
 * @inheritdoc
 */
public function behaviors()
{
    return [
        'access' => [
            'class' => AccessControl::className(),
            'rules' => [
                [
                    'allow' => true,
                    'actions' => ['index'],
                    'roles' => ['@'],
                ],
                [
                    'allow' => !Yii::$app->user->isGuest && Yii::$app->user->identity->isAdmin(),
                    'actions' => ['view', 'create', 'update', 'delete'],
                ],
            ],
        ],
    ];
}

然后将User模型添加到isAdmin(),其中为您的管理员用户返回true,为其他人返回false。就个人而言,我使用:

public function isAdmin() {
    return Self::ROLE_ADMIN === $this->role;
}

不可否认,这不是“按书”。但它简单,快速,有效。

答案 4 :(得分:0)

$user_id = \Yii::$app->user->id;

$auth = new DbManager;
$auth->init();
$role = $auth->createRole('editor');
$auth->add($role);

$auth->assign($role, $user_id);
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