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主页 / 问题 / 1258164
Accepted
Elizaveta
Elizaveta
Asked:2022-03-19 23:10:23 +0000 UTC2022-03-19 23:10:23 +0000 UTC 2022-03-19 23:10:23 +0000 UTC

如何处理自定义 JwtAuthException Spring 安全性

  • 772

我有 JwtTokenProvider:

@Component
@PropertySource("classpath:jwt.properties")
@RequiredArgsConstructor
public class JwtTokenProvider {
//some code ...
   public boolean validateToken(String token) {
        try {
            Jws<Claims> claimsJws = Jwts.parser()
                    .setSigningKey(secretKey)
                    .parseClaimsJws(token);
            return !claimsJws.getBody().getExpiration().before(new Date());
        } catch (JwtException | IllegalArgumentException e) {
            throw new JwtAuthenticationException("JWT token expired or invalid!");
        }
    }
}

这是一个正常组件,如果令牌“腐烂”,结果会引发异常

很明显,如果您使用损坏的令牌联系服务器,它将抛出异常 - 但会出现 500 错误。如何捕获它以便返回 401?而且只有这个例外,403我需要保存

以下是一些相关的类:

  1. 筛选
@RequiredArgsConstructor
public class JwtTokenFilter extends GenericFilterBean {

    private final JwtTokenProvider jwtTokenProvider;

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
        String token = jwtTokenProvider.retrieveToken((HttpServletRequest) servletRequest);

        if (token != null && jwtTokenProvider.validateToken(token)) {
            Authentication authentication = jwtTokenProvider.getAuthentication(token);

            if (authentication != null) {
                SecurityContextHolder.getContext().setAuthentication(authentication);
            }
        }
        filterChain.doFilter(servletRequest, servletResponse);
    }
}

2.配置

@RequiredArgsConstructor
public class JwtConfigurer extends SecurityConfigurerAdapter<DefaultSecurityFilterChain, HttpSecurity> {
    private final JwtTokenProvider provider;

    @Override
    public void configure(HttpSecurity builder) throws Exception {
        JwtTokenFilter filter = new JwtTokenFilter(provider);
        builder.addFilterBefore(filter, UsernamePasswordAuthenticationFilter.class);
    }
}
==========================
@Configuration
@EnableWebSecurity
@ComponentScan(basePackageClasses = {SecurityPackageMarker.class})
@EnableGlobalMethodSecurity(prePostEnabled = true)
@RequiredArgsConstructor
public class SecurityConfig extends WebSecurityConfigurerAdapter {

    private final JwtTokenProvider provider;

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
                .cors()
                .and()
                .addFilterBefore(new CorsFilter(corsConfigurationSource()), SessionManagementFilter.class)
                .httpBasic().disable()
                .csrf().disable()
                .sessionManagement().sessionCreationPolicy(SessionCreationPolicy.STATELESS)
                .and()
                .authorizeRequests()
                .antMatchers("/hello", "/login", "/register").permitAll()
                .anyRequest().authenticated()
                .and()
                .apply(new JwtConfigurer(provider));
    }

    @Bean
    @Override
    public AuthenticationManager authenticationManagerBean() throws Exception {
        return super.authenticationManagerBean();
    }

    @Bean
    public CorsConfigurationSource corsConfigurationSource() {
        UrlBasedCorsConfigurationSource source = new UrlBasedCorsConfigurationSource();
        CorsConfiguration config = new CorsConfiguration();
        config.addAllowedMethod("*");
        source.registerCorsConfiguration("/**", config.applyPermitDefaultValues());
        return source;
    }
}
  1. 例外:
public class JwtAuthenticationException extends AuthenticationException {
    public JwtAuthenticationException(String explanation) {
        super(explanation);
    }
}

4.控制器:

@PostMapping("/login")
    @ResponseStatus(HttpStatus.OK)
    public Properties login(@RequestBody AuthenticationInDto dto) {
        try {
            String username = dto.getUsername();
            String password= dto.getPassword();
            authenticationManager.authenticate(new UsernamePasswordAuthenticationToken(username, password));
            Optional<User> user = userService.findByUsername(username);

            if (!user.isPresent()) {
                throw new UsernameNotFoundException("User with username: " + username + " not found");
            }

            String token = provider.createToken(username, user.get().getRole());

            Properties response = new Properties();
            response.setProperty("jwtToken", token);

            return response;
        } catch (AuthenticationException e) {
            throw new BadCredentialsException(e.getMessage());
        }
    }

但是,尽管如此,它并没有捕获我的异常(不转发它BadCredentialsException) - 服务器崩溃并出现错误 500,即使我有一个处理程序:

@RestControllerAdvice
public class ControllersExceptionHandler extends ResponseEntityExceptionHandler {

    @ExceptionHandler(BadCredentialsException.class)
    @ResponseStatus(HttpStatus.UNAUTHORIZED)
    public ErrorDto handleBadCredentials(BadCredentialsException ex) {
        return new ErrorDto(HttpStatus.UNAUTHORIZED.value(),
                            ex.getMessage());
    }

    @ExceptionHandler(RegistrationException.class)
    @ResponseStatus(HttpStatus.BAD_REQUEST)
    public ErrorDto handleRegistrationException(RegistrationException ex) {
        return new ErrorDto(HttpStatus.BAD_REQUEST.value(),
                            ex.getMessage());
    }
}
java
  • 1 1 个回答
  • 10 Views

1 个回答

  • Voted
  1. Best Answer
    lampa
    2022-03-22T20:37:48Z2022-03-22T20:37:48Z

    我已经简化为过滤器中的任何错误:

    @Component
    public class TokenFilter extends GenericFilterBean {
        @Override
        public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
            String token = servletRequest.getParameter("token");
    
            if (token == null || !token.equals("123456")) {
                HttpServletResponse httpServletResponse = (HttpServletResponse) servletResponse;
                httpServletResponse.sendError(HttpStatus.UNAUTHORIZED.value());
            }
    
            filterChain.doFilter(servletRequest, servletResponse);
        }
    }
    

    在您的情况下,捕获过滤器中的错误并发送sendError(). 在sendError()第二个参数中,您可以添加错误描述。

    validateToken()如果在方法中处理了异常,并且接口GenericFilterBean将方法设置为仅 2 个可能的异常,则仍然不清楚您是如何设法编译的:IOException和ServletException.

    从理论上讲,@RestControllerAdvice它不会通过捕获错误来工作,因为。servlet -> 过滤器前 -> spring mvc -> 过滤器后。

    顺便说一句,@Aspect它工作得很好:

    @Aspect
    @Component
    public class ExceptionHandlerAspect {
    
        @Before(value = "within(@org.springframework.web.bind.annotation.RestController *)")
        public void before(JoinPoint joinPoint) {
            HttpServletRequest request = ((ServletRequestAttributes) RequestContextHolder.currentRequestAttributes()).getRequest();
            String token = request.getParameter("token");
    
            if (token == null || !token.equals("123456")) {
                throw new BadCredentialsException(HttpStatus.UNAUTHORIZED, "bad token");
            }
        }
    }
    
    • 1

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