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使用自定义UserDetailsS​​ervice进行Spring Boot

[英]Spring Boot with custom UserDetailsService

将我自己的UserDetailsS​​ervice(使用Spring Data JPA)的自定义实现添加到Spring Boot应用程序的正确方法是什么?

public class DatabaseUserDetailsService implements UserDetailsService {

    @Inject
    private UserAccountService userAccountService;

    @Override
    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
        User user = userAccountService.getUserByEmail(username);
        return new MyUserDetails(user);
    }

}


public interface UserRepository extends JpaRepository<User, Long>, JpaSpecificationExecutor<User> {

    public User findByEmail(String email);

}



@Service
public class UserAccountService {

    @Inject
    protected UserRepository userRepository;

    public User getUserByEmail(String email) {
        return userRepository.findByEmail(email);
    }

}


@Configuration
@ComponentScan
@EnableAutoConfiguration
@EnableGlobalMethodSecurity(prePostEnabled = true)
@EnableTransactionManagement
@EnableJpaRepositories(basePackages = "com.sample")
@EntityScan(basePackages = { "com.sample" })
@EnableJpaAuditing(auditorAwareRef = "auditorProvider")
public class Application {

    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }

    ...

    @Order(SecurityProperties.ACCESS_OVERRIDE_ORDER)
    protected static class ApplicationSecurity extends WebSecurityConfigurerAdapter {

        @Override
        protected void configure(HttpSecurity http) throws Exception {
            http
                .authorizeRequests()
                    .antMatchers("/").hasRole("USER")
                    .and()
                .formLogin()
                    .loginPage("/login")
                    .permitAll()
                    .and()
                .logout()
                    .permitAll();
        }


    }

    @Order(Ordered.HIGHEST_PRECEDENCE + 10)
    protected static class AuthenticationSecurity extends GlobalAuthenticationConfigurerAdapter {

        @Inject
        private UserAccountService userAccountService;

        @Override
        public void init(AuthenticationManagerBuilder auth) throws Exception {
            auth.userDetailsService(userDetailsService());
        }

        @Bean
        public UserDetailsService userDetailsService() {
            return new DatabaseUserDetailsService();
        }

    }

}


@Entity
public class User extends AbstractPersistable<Long> {

    @ManyToMany
    private List<Role> roles = new ArrayList<Role>();

    // getter, setter

}


@Entity
public class Role extends AbstractPersistable<Long> {

    @Column(nullable = false)
    private String authority;

    // getter, setter

}

我无法启动app beacouse我得到(完全例外这里http://pastebin.com/gM804mvQ

Caused by: org.hibernate.AnnotationException: Use of @OneToMany or @ManyToMany targeting an unmapped class: com.sample.model.User.roles[com.sample.model.Role]
    at org.hibernate.cfg.annotations.CollectionBinder.bindManyToManySecondPass(CollectionBinder.java:1134)

当我使用auth.jdbcAuthentication().dataSource(dataSource).usersByUsernameQuery("...).authoritiesByUsernameQuery("...")配置我的ApplicationSecurity ,一切正常,包括JPA和Spring Data存储库。

您的应用似乎对我@Configuration (一旦我将@Configuration添加到AuthenticationSecurity )。 这是JPA UserDetailsService的一个简单应用程序的另一个工作示例,以防它有用: https//github.com/scratches/jpa-method-security-sample

您还可以关注此博客以实现自定义用户详细信息服务。

此示例显示如何将bean发送到userdetails服务以进行注入。

  1. 在WebSecurityConfigurer中自动装配存储库
  2. 通过参数化构造函数将此bean作为参数发送到用户详细信息服务。
  3. 为此分配一个私有成员并用于从数据库加载用户。

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