[英]Spring Boot with custom UserDetailsService
What is the correct way to add my custom implementation of UserDetailsService (which uses Spring Data JPA) to Spring Boot app? 将我自己的UserDetailsService(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)
When I configure my ApplicationSecurity
with auth.jdbcAuthentication().dataSource(dataSource).usersByUsernameQuery("...).authoritiesByUsernameQuery("...")
everything is working including JPA and Spring Data repositories. 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。
Your app seems to work for me (once I added @Configuration
to the AuthenticationSecurity
). 您的应用似乎对我@Configuration
(一旦我将@Configuration
添加到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
You can also follow this blog to implement custom user details service. 您还可以关注此博客以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. 此示例显示如何将bean发送到userdetails服务以进行注入。
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