Spring Security中如何通过用户ID查询用户

如何解决Spring Security中如何通过用户ID查询用户

我有一个带有Spring Security的Spring Boot应用程序，其中扩展了org.springframework.security.core.userdetails.User类以创建自定义用户对象。我是硬编码用户，因为我只需要几个来测试一些功能。我根据用户ID或电子邮件ID查询并获取某个用户，这是我添加的自定义变量？有没有一种方法可以执行此操作而无需切换到jdbc身份验证等，因为我觉得它可能会使它变得过于复杂了？

我的自定义用户类别：

import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.userdetails.User;
import java.util.Collection;
import java.util.List;

public class Guest extends User {
    public Guest(String username,String password,boolean enabled,boolean accountNonExpired,boolean credentialsNonExpired,boolean accountNonLocked,Collection<? extends GrantedAuthority> authorities) {
        super(username,password,enabled,accountNonExpired,credentialsNonExpired,accountNonLocked,authorities);
    }
    private long id;
    private String firstName;
    private String lastName;
    private String emailAddress;
    private String phoneNumber;
    
    //some getters and setters

    public Guest(long id,String firstName,String lastName,String emailAddress,String phoneNumber,String username,List<GrantedAuthority> authorities,boolean accountNonLocked) {


        super(username,authorities) ;
        this.id=id;
        this.firstName=firstName;
        this.lastName=lastName;
        this.emailAddress=emailAddress;
        this.phoneNumber=phoneNumber;
    }
}

我的新userdetailsS​​ervice类：

@Service("userDetailsService")
public class MyUserDetailsService implements UserDetailsService {
    private static List<Guest> users = new ArrayList();

    public MyUserDetailsService() { 
    //hard-coding users
}
    @Override
    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
    //building user here
}

SecurityConfig.java

@Configuration
@EnableWebSecurity
public class ApplicationSecurityConfiguration extends WebSecurityConfigurerAdapter{
    
    @Autowired
    @Qualifier("userDetailsService")
    private UserDetailsService userDetailsService;

    @Bean
    public BCryptPasswordEncoder passwordEncoder() {
        return new BCryptPasswordEncoder();
    };

    @Override
    protected void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth.userDetailsService(userDetailsService).passwordEncoder(passwordEncoder());
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
       //authorization logic here        
    }

}

解决方法

您可以像下面的示例那样使用硬编码的值来做..如果需要进行模拟实现

public UserDetails loadUserByUsername(String userName){
        
//use your custom 'List<Guest> users' hardcoded values to create User object

User user = new User(username,password,enabled,accountNonExpired,credentialsNonExpired,accountNonLocked,authorities);
return user;
}

注意：-loadUserByUsername（String username）..这里的userName可以是userToken，其中包含您想要作为参考的任何信息。 （通常用于查询db来获取usrname并通过它的查询）..在这种情况下，可以说id ..一旦获得上面的id（作为userToken / userName），就可以与您的来宾列表进行比较以加载User对象