带有自定义 UserDetailsS​​ervice 的 Spring Boot

将我的 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?

What is the correct way to add my custom implementation of UserDetailsService (which uses Spring Data JPA) to Spring Boot app?

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

}

我无法启动应用程序,因为我得到了(此处完全例外 http://pastebin.com/gM804mvQ)

I cannot start app beacouse I get (full exception here 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.

When I configure my ApplicationSecurity with auth.jdbcAuthentication().dataSource(dataSource).usersByUsernameQuery("...).authoritiesByUsernameQuery("...") everything is working including JPA and Spring Data repositories.

推荐答案

您的应用似乎对我有用(一旦我将 @Configuration 添加到 AuthenticationSecurity).这是另一个使用 JPA UserDetailsS​​ervice 的简单应用程序的工作示例,以防万一:https://github.com/scratches/jpa-method-security-sample

Your app seems to work for me (once I added @Configuration to the AuthenticationSecurity). Here's another working sample of a simple app with JPA UserDetailsService in case it helps: https://github.com/scratches/jpa-method-security-sample

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