Spring Security 3 database authentication with Hibernate

Question

I need to authenticate users from database, Spring Security documents don't tell how to authenticate with hibernate. Is that possible and how can I do that?

Answer 1

You have to make you're own custom authentication-provider.

Example code:

Service to load Users from Hibernate:

import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;    

@Service("userDetailsService") 
public class UserDetailsServiceImpl implements UserDetailsService {

  @Autowired private UserDao dao;
  @Autowired private Assembler assembler;

  @Transactional(readOnly = true)
  public UserDetails loadUserByUsername(String username)
      throws UsernameNotFoundException, DataAccessException {

    UserDetails userDetails = null;
    UserEntity userEntity = dao.findByName(username);
    if (user == null)
      throw new UsernameNotFoundException("user not found");

    return = assembler.buildUserFromUserEntity(userEntity);
  }
}

Service to conver you're entity to a spring user object:

import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.GrantedAuthorityImpl;
import org.springframework.security.core.userdetails.User;

@Service("assembler")
public class Assembler {

  @Transactional(readOnly = true)
  User buildUserFromUserEntity(UserEntity userEntity) {

    String username = userEntity.getName();
    String password = userEntity.getPassword();
    boolean enabled = userEntity.isActive();
    boolean accountNonExpired = userEntity.isActive();
    boolean credentialsNonExpired = userEntity.isActive();
    boolean accountNonLocked = userEntity.isActive();

    Collection<GrantedAuthority> authorities = new ArrayList<GrantedAuthority>();
    for (SecurityRoleEntity role : userEntity.getRoles()) {
      authorities.add(new GrantedAuthorityImpl(role.getRoleName()));
    }

    User user = new User(username, password, enabled,
      accountNonExpired, credentialsNonExpired, accountNonLocked, authorities, id);
    return user;
  }
}

The namespace-based application-context-security.xml would look something like:

<http>
  <intercept-url pattern="/login.do*" filters="none"/>
  <intercept-url pattern="/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
  <form-login login-page="/login.do"
              authentication-failure-url="/login.do?error=failed"
              login-processing-url="/login-please.do" />
  <logout logout-url="/logoff-please.do"
          logout-success-url="/logoff.html" />
</http>

<beans:bean id="daoAuthenticationProvider"
 class="org.springframework.security.authentication.dao.DaoAuthenticationProvider">
  <beans:property name="userDetailsService" ref="userDetailsService"/>
</beans:bean>

<beans:bean id="authenticationManager"
    class="org.springframework.security.authentication.ProviderManager">
  <beans:property name="providers">
    <beans:list>
      <beans:ref local="daoAuthenticationProvider" />
    </beans:list>
  </beans:property>
</beans:bean>

<authentication-manager>
  <authentication-provider user-service-ref="userDetailsService">
    <password-encoder hash="md5"/>
  </authentication-provider>
</authentication-manager>

Answer 2

Hi guys im currently using Spring Security 2.x. Can i just check, are UserEntity and SecurityRoleEntity valid types? because it seems that they are not valid on my implementation. please advice!

Answer 3

When using the code above,I got error: Caused by: org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'userDetailsService' is defined at org.springframework.beans.factory.support.DefaultListableBeanFactory.getBeanDefinition(DefaultListableBeanFactory.java:510) at org.springframework.beans.factory.support.AbstractBeanFactory.getMergedLocalBeanDefinition(AbstractBeanFactory.java:1056) at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:274) at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:190) at org.springframework.beans.factory.support.BeanDefinitionValueResolver.resolveReference(BeanDefinitionValueResolver.java:322) ... 56 more

Answer 4

Thanks for that. I have used with jdbctemplate.