Java 带有自定义 UserDetailsS​​ervice 的 Spring Boot

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时间:2020-08-14 14:27:51  来源:igfitidea点击:

Spring Boot with custom UserDetailsService

javaspringspring-securityspring-dataspring-boot

提问by igo

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

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?

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

}

I cannot start app beacouse I get (full exception here http://pastebin.com/gM804mvQ)

我无法启动应用程序,因为我得到了(这里完全例外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)

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

当我配置我ApplicationSecurityauth.jdbcAuthentication().dataSource(dataSource).usersByUsernameQuery("...).authoritiesByUsernameQuery("...")一切时,包括 JPA 和 Spring Data 存储库。

采纳答案by Dave Syer

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

您的应用程序似乎对我有用(一旦我添加@ConfigurationAuthenticationSecurity)。这是使用 JPA 的简单应用程序的另一个工作示例,UserDetailsService以防万一:https: //github.com/scratches/jpa-method-security-sample

回答by Ekansh Rastogi

You can also follow this blogto implement custom user details service.

您也可以关注此博客来实现自定义用户详细信息服务。

This example shows how you can send bean to userdetails service for injection.

此示例展示了如何将 bean 发送到 userdetails 服务进行注入。

  1. Autowire the Repository in the WebSecurityConfigurer
  2. Send this bean as a parameter to the user details service by a parameterized constructor.
  3. assign this a private member and use to load users from database.
  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. 为其分配一个私有成员并用于从数据库加载用户。