6

我在我的 Grails 应用程序中使用 spring 安全插件,我希望注册的用户使用注册过程中提供的用户名或电子邮件地址登录应用程序。(就像 fb 一样)

我已经查看了网络并关注 Grails 文档 ::: http://grails-plugins.github.io/grails-spring-security-core/docs/manual/guide/single.html#4.1%20Person%20Class

但找不到任何解决方案。

任何帮助将不胜感激。

提前致谢

4

2 回答 2

8

从 spring-security 的角度来看,你应该做的是实现你自己的org.springframework.security.core.userdetails.UserDetailsS​​ervice

public class MyUserServiceImpl implements UserDetailsService{

    @Autowired
    private AdminUserDao dao;

    @Override
    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException, DataAccessException {

    ... Here comes your logic to get your the user based on email or userid ...

    }
}

在您的春季配置中,您必须引用您的 UserDetails 类:

<authentication-manager>
    <authentication-provider user-service-ref="myUserServiceImpl" />
</authentication-manager>

hth

于 2013-09-29T07:58:26.083 回答
0

我不认为这有什么最近的事情。在更高版本的 spring security / grails 3.2.8 中。这是我的:

这是在服务中创建的服务。用户电子邮件位于用户之外的一个单独的类中,因此attributes.email检查,否则你可以这样做User.findByUsernameOrEmail(username,username)

所以我有 UserEmailUserService 作为服务:

package com.app.users

import grails.plugin.springsecurity.userdetails.GormUserDetailsService
import grails.transaction.Transactional
import org.springframework.security.authentication.UsernamePasswordAuthenticationToken
import org.springframework.security.core.Authentication
import org.springframework.security.core.GrantedAuthority
import org.springframework.security.core.authority.SimpleGrantedAuthority
import org.springframework.security.core.context.SecurityContextHolder
import org.springframework.security.core.userdetails.UserDetails
import org.springframework.security.core.userdetails.UsernameNotFoundException

class UserEmailUserService extends GormUserDetailsService{

    UserDetails loadUserByUsername(String username, boolean loadRoles)
            throws UsernameNotFoundException {
        return loadUserByUsername(username)
    }

    @Transactional
    UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
        //enable login with either username or password
        User user = User.find {
            username == username || attributes.email == username
        }
        if (!user) throw new UsernameNotFoundException('User not found', username)
        UserDetails userDetails = new org.springframework.security.core.userdetails.User(user.username, user.getPassword(),
                user.enabled, !user.accountExpired, !user.passwordExpired, !user.accountLocked, getAuthorities(user.roles))
        Authentication authentication = new UsernamePasswordAuthenticationToken(userDetails, null, userDetails.getAuthorities())
        SecurityContextHolder.getContext().setAuthentication(authentication);
        return userDetails
    }

    public static List<GrantedAuthority> getAuthorities(Set<Role> roles) {
       List<GrantedAuthority> authorities = new ArrayList<GrantedAuthority>()
       roles?.each { role ->
          authorities.add(new SimpleGrantedAuthority(role.authority))
       }
       return authorities
   }
}

然后在conf/spring/resources.groovy

import com.app.users.UserEmailUserService

// Place your Spring DSL code here
beans = {
      userDetailsService(UserEmailUserService){
            grailsApplication = ref('grailsApplication')
        }
}

请注意我调用user.roles了这是我在用户类中进行的自定义调用。因为我有团体角色等:

Set<Role> getRoles() {
    UserRole.findAllByUser(this)*.role
}
于 2017-04-28T20:06:43.263 回答