编写自定义 Shiro 领域

Writing custom Shiro realm

提问人:smeeb 提问时间:11/21/2014 最后编辑:cheb1k4smeeb 更新时间:11/18/2023 访问量:10410

问:

我正在构建自己的子类,并且很难将其连接到我的.AuthorizingRealmSecurityManager

我境界的本质:

public class MyRealm extends AuthorizingRealm { 
    protected AuthenticationInfo doGetAuthenticationInfo(AuthenticationToken token) throws AuthenticationException { 
        try { 
            // My custom logic here 

        } catch(Throwable t) { 
            System.out.println(t.getMessage()); 
        } 
        SimpleAuthenticationInfo authn = new SimpleAuthenticationInfo(new MyUser(), "somePassword");
        return authn;
    } 

    protected AuthorizationInfo doGetAuthorizationInfo(PrincipalCollection principals) { 
        try { 
            // My custom logic here 
        } catch(Throwable t) { 
            System.out.println(t.getMessage()); 
        }
        return new SimpleAuthorizationInfo();
    } 
}

然后在我的“shiro.ini”中:

# ======================= 
# Shiro INI configuration 
# ======================= 
[main] 
myRealm = com.me.myapp.security.MyRealm 

然后在我的 Driver 类/main 方法(我用于测试)中:

public class Driver { 
    public static void main(String[] args) { 
        Driver d = new Driver(); 
        d.test(); 
    } 

    public void test() { 
        Factory<SecurityManager> factory = new IniSecurityManagerFactory("classpath:shiro.ini"); 
        SecurityManager securityManager = factory.getInstance(); 
        SecurityUtils.setSecurityManager(securityManager); 

        UsernamePasswordToken token = new UsernamePasswordToken("", ""); 
        token.setRememberMe(true); 

        System.out.println("Shiro props:"); 
        System.out.println(securityManager.getProperties()); 

        Subject currentUser = SecurityUtils.getSubject() 

        try { 
            currentUser.login(token) 

            println "I think this worked!" 
        } catch (UnknownAccountException uae) { 
            println "Exception: ${uae}" 
        } catch (IncorrectCredentialsException ice) { 
            println "Exception: ${ice}" 
        } catch (LockedAccountException lae) { 
            println "Exception: ${lae}" 
        } catch (ExcessiveAttemptsException eae) { 
            println "Exception: ${eae}" 
        } catch (AuthenticationException ae) { 
            println "Exception: ${ae}" 
        } 
    } 
} 

当我运行它时,我得到:

Shiro props: 
[class:class org.apache.shiro.mgt.DefaultSecurityManager, cacheManager:null, subjectFactory:org.apache.shiro.mgt.DefaultSubjectFactory@6a2b8b42, authorizer:org.apache.shiro.authz.ModularRealmAuthorizer@50c3d082, realms:[com.me.myapp.security.MyRealm@67ae303a], subjectDAO:org.apache.shiro.mgt.DefaultSubjectDAO@5ce06503, rememberMeManager:null, authenticator:org.apache.shiro.authc.pam.ModularRealmAuthenticator@1007d798, sessionManager:org.apache.shiro.session.mgt.DefaultSessionManager@72db4460] 
Exception: org.apache.shiro.authc.AuthenticationException: Authentication failed for token submission [org.apache.shiro.authc.UsernamePasswordToken - , rememberMe=true].  Possible unexpected error? (Typical or expected login exceptions should extend from AuthenticationException). 

所以它看起来像在读取我的 shiro.ini,因为它选择了正确的领域,但除了存根应该进行身份验证的虚拟用户之外,无论提供的用户名/密码如何,它都没有做任何事情。关于我哪里出了问题的任何想法?MyRealm

安全 认证 授权 Shiro

评论


答:

0赞 Luca Rasconi 12/2/2014 #1

将此添加到您的 shiro.ini:
然后在 Driver 类中
securityManager.realms = $myRealm

UsernamePasswordToken token = new UsernamePasswordToken("", "somePassword"); 

而不是一个空的密码。

我认为这奏效了!

评论

0赞 smeeb 12/28/2014
谢谢@Luca Rasconi,但是您的建议不会改变任何事情(与我上面描述的行为相同)。还有其他想法/想法吗?再次感谢!
0赞 Cam 1/4/2015 #2

我自己没有这样做过,但您可以尝试以下几件事:

  1. 如果不需要授权逻辑,请考虑将 AuthenticatingRealm 而不是 AuthorizingRealm 子类化

  2. 在方法 doGetAuthenticationInfo 中,请考虑使用以下代码:

    SimpleAuthenticationInfo authn = new SimpleAuthenticationInfo(token.getPrincipal(), token.getCredentials(), “myRealm”);

0赞 Akshay Anurag 6/26/2019 #3

你似乎已经正确地创建了一个领域,但没有告诉它是一个要使用的领域。这样,它只是在 的部分中创建的另一个对象。SecurityManagermainshiro.ini

要告诉 Shiro's 它需要用作 Realm,您需要将其添加到您的:SecurityManagermyRealmshiro.ini

securityManager.realms = $myRealm