java - Spring 启动 : Get current user's username

Question

我正在尝试使用 Spring 的安全性获取当前登录的用户名，但 Principal 对象返回 null。

这是我的 REST Controller 方法:

@RequestMapping("/getCurrentUser")
public User getCurrentUser(Principal principal) {

    String username = principal.getName();
    User user = new User();
    if (null != username) {
        user = userService.findByUsername(username);
    }

    return user;
}

注意:我正在运行 Spring boot 1.5.13 和 spring security 4.2.6

• 这是我的安全配置类:

  @Configuration
  
  @EnableWebSecurity
  
  public class SecurityConfig extends WebSecurityConfigurerAdapter{
  
      @Autowired
      private Environment env;
  
      @Autowired
      private UserSecurityService userSecurityService;
  
      private BCryptPasswordEncoder passwordEncoder() {
          return SecurityUtility.passwordEncoder();
      }
  
      private static final String[] PUBLIC_MATCHERS = {
              "/css/**",
              "/js/**",
              "/image/**",
              "/book/**",
              "/user/**"
      };
  
      @Override
      protected void configure(HttpSecurity http) throws Exception {
          http.csrf().disable().cors().disable().httpBasic().and().authorizeRequests()
          .antMatchers(PUBLIC_MATCHERS).permitAll().anyRequest().authenticated();
      }
  
      @Autowired
      public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
          auth.userDetailsService(userSecurityService).passwordEncoder(passwordEncoder());
      }
  
      @Bean
      public HttpSessionStrategy httpSessionStrategy() {
          return new HeaderHttpSessionStrategy();
      }
  }
  

• 这是我的用户安全服务类:

  @Service
  
  public class UserSecurityService implements UserDetailsService {
  
  private static final Logger LOG = LoggerFactory.getLogger(UserSecurityService.class);
  
      @Autowired 
      private UserRepository userRepository;
  
      @Override
      public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
          User user = userRepository.findByUsername(username);
          if(null == user) {
              LOG.warn("Username {} not found", username);
              throw new UsernameNotFoundException("Username "+username+" not found");
          }
          return user;
      }
  }
  

• 这是我的用户类:

  @Entity
  
  public class User implements UserDetails, Serializable {
  
      private static final long serialVersionUID = 902783495L;
  
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      @Column(name="Id", nullable=false, updatable = false)
      private Long id;
  
      private String username;
      private String password;
      private String firstName;
      private String lastName;
  
      private String email;
      private String phone;
      private boolean enabled = true;
  
      @OneToMany(mappedBy = "user", cascade=CascadeType.ALL, fetch = FetchType.EAGER)
      @JsonIgnore
      private Set<UserRole> userRoles = new HashSet<>();
  }
  

Answer 1

有多种方法可以做到这一点。

使用SecurityContextHolder

Authentication auth = SecurityContextHolder.getContext().getAuthentication();
String username = auth.getName();

从 Controller 使用Principal

@RequestMapping(value = "/myusername", method = RequestMethod.GET)
@ResponseBody
public String currentUserName(Principal principal) {
    return principal.getName();
}

来自 HttpServletRequest

@RequestMapping(value = "/myusername", method = RequestMethod.GET)
@ResponseBody
public String getUsername(HttpServletRequest req) {
    return req.getUserPrincipal.getName();
}