从 Spring 上的当前会话获取用户信息？

Question

我一直在尝试从当前会话中获取用户信息以执行一些类似 findByUsername 的操作。我已经尝试过@AuthenticationPrinciple，但即使我用我的 UserDetails 实现来提供它，它也只是 returns null。我还尝试了 returns anonymousUser(?) 的 SecurityContextHolder 方法。无论哪种方式都没有得到想要的结果。尝试了到目前为止我能在互联网上找到的所有解决方案，但没有成功。控制器；

@Controller
public class Home {

    EntryService entryService;

    public Home(EntryService entryService) {
        this.entryService = entryService;
    }


    @GetMapping("/Home")
    public String registration(Entry entry, Model model) {
        //See what it returns
        System.out.println(getUsername());
        List<Entry> entries = new ArrayList<>(entryService.getAllEntries());
        model.addAttribute("entryList", entries);
        model.addAttribute("entry", entry);

        return "/home";
    }


    public String getUsername() {
        SecurityContext context = SecurityContextHolder.getContext();
        Authentication authentication = context.getAuthentication();
        if (authentication == null)
            return null;
        Object principal = authentication.getPrincipal();
        if (principal instanceof UserDetails) {
            return ((UserDetails) principal).getUsername();
        } else {
            return principal.toString();
        }
    }

}

安全；

@Configuration

public class SecurityConfig extends WebSecurityConfigurerAdapter {

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

    @Autowired
    public DetailsService detailsService() {
        return new DetailsService();
    }

    protected void configure(HttpSecurity http) throws Exception {

        http.
                authorizeRequests().
                antMatchers("/register").
                permitAll().
                antMatchers("/home").
                hasRole("USER").
                and().
                csrf().
                disable().
                formLogin().
                loginPage("/").
                permitAll().
                passwordParameter("password").
                usernameParameter("username").
                defaultSuccessUrl("/home").
                failureUrl("/error").
                and().
                logout().
                logoutUrl("/logout");
    }

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

}

用户详细信息；

public class UserDetail implements UserDetails {

    private final String username;
    private final String password;
    private final boolean active;
    private final List<GrantedAuthority> roles;

    public UserDetail(User user) {
        this.username = user.getUserName();
        this.password = user.getPassword();
        this.active = user.getActive();
        this.roles = Arrays.stream(user.getRole().toString().split(",")).
                map(SimpleGrantedAuthority::new).
                collect(Collectors.toList());
    }

    @Override
    public Collection<? extends GrantedAuthority> getAuthorities() {
        return roles;
    }

    @Override
    public String getPassword() {
        return password;
    }

    @Override
    public String getUsername() {
        return username;
    }

    @Override
    public boolean isAccountNonExpired() {
        return true;
    }

    @Override
    public boolean isAccountNonLocked() {
        return true;
    }

    @Override
    public boolean isCredentialsNonExpired() {
        return true;
    }

    @Override
    public boolean isEnabled() {
        return active;
    }
}

和 UserDetailsService；

@Service
public class DetailsService implements UserDetailsService {

    @Autowired
    UserRepository userRepository;

    @Override
    public UserDetails loadUserByUsername(String s) throws UsernameNotFoundException {
        Optional<User> user = userRepository.findByUserName(s);
        user.orElseThrow(() -> new UsernameNotFoundException("User not found"));
        return user.map(UserDetail::new).get();
    }
}

顺便说一句，使用基于 JPA 的身份验证，它可以正常工作。

Answer 1

您在安全上下文中获得 anonymousUser 的唯一原因是您未通过身份验证。尝试在您的 SecurityConfig 中的 hasRole("USER"). 之后添加 .anyRequest().authenticated()，然后您应该会在 SecurityContextHolder.getContext().getAuthentication() 中看到主体。这将继续使用您指定为 permitAll().

的方法

此外，只是一个观察，但是您配置中的 url 匹配器在 /home 上并且您的控制器指定了 /Home.

的 GetMapping

从 Spring 上的当前会话获取用户信息？

Getting user information from the current session on Spring?

java

session

spring

spring-security

user-data