带有自定义UserDetailsService的Spring Boot [英] Spring Boot with custom UserDetailsService
问题描述
将我的UserDetailsService的自定义实现(使用Spring Data JPA)添加到Spring Boot应用程序中的正确方法是什么?
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
}
我无法启动我的应用程序,因为我得到了(此处为 http://pastebin.com/gM804mvQ 的完全例外)
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)
当我用auth.jdbcAuthentication().dataSource(dataSource).usersByUsernameQuery("...).authoritiesByUsernameQuery("...")
配置ApplicationSecurity
时,一切都正常,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(一旦我在AuthenticationSecurity
中添加了@Configuration
).如果有帮助,这是使用JPA UserDetailsService
的简单应用程序的另一个工作示例: https://github.com/划痕/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
这篇关于带有自定义UserDetailsService的Spring Boot的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!