使用自定義UserDetailsS​​ervice進行Spring Boot

Question

將我自己的UserDetailsS​​ervice（使用Spring Data JPA）的自定義實現添加到Spring Boot應用程序的正確方法是什么？

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

}

我無法啟動app beacouse我得到（完全例外這里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)

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

Answer 1

您的應用似乎對我@Configuration （一旦我將@Configuration添加到AuthenticationSecurity ）。 這是JPA UserDetailsService的一個簡單應用程序的另一個工作示例，以防它有用： https ： //github.com/scratches/jpa-method-security-sample

Answer 2

您還可以關注此博客以實現自定義用戶詳細信息服務。

此示例顯示如何將bean發送到userdetails服務以進行注入。

1. 在WebSecurityConfigurer中自動裝配存儲庫
2. 通過參數化構造函數將此bean作為參數發送到用戶詳細信息服務。
3. 為此分配一個私有成員並用於從數據庫加載用戶。