java - Spring Security OAuth2 舞蹈和获取参数

Question

在我的 Java Spring 应用程序中，我通过外部 OAuth2 提供程序实现了 OAuth2 用户授权。

在我的本地主机上，为了通过这个外部 OAuth2 提供商对用户进行身份验证，我需要访问以下网址:https://127.0.0.1:8443/login/ok在 OAuth2 dance 之后，我可以让这个用户通过身份验证。到目前为止一切正常。

但是当我的登录 url 中有一些请求参数时，例如 uid 和级别:

https://127.0.0.1:8443/login/ok?uid=45134132&level=3

在 OAuth2 舞蹈之后我被重定向到 https://127.0.0.1:8443/并丢失这些参数。

在我的 Chrome 网络面板中，我可以看到以下一组调用:

1. > https://127.0.0.1:8443/login/ok?uid=45134132&level=3
2. > https://connect.ok.ru/oauth/authorize?redirect_uri=https://127.0.0.1:8443/login/ok?uid%3D45134132%26level%3D3&response_type=code&state=AKakq ....
3. > https://127.0.0.1:8443/login/ok?uid=45134132&level=3&code= ....
4. > https://127.0.0.1:8443/

所以我在第 3 步之后丢失了这些参数。

是否可以配置 Spring Security + OAuth2 以将这些参数也传递给步骤 #4？

这是我的配置(这是基于此答案的解决方案 Spring Security - Retaining URL parameters on redirect to login )但它不起作用(未调用AuthenticationProcessingFilterEntryPoint .commence 方法):

    @Override
    public void configure(HttpSecurity http) throws Exception {
        // @formatter:off   
        http
        .headers().frameOptions().disable()
        .and().logout()
        .and().antMatcher("/**").authorizeRequests()
            .antMatchers("/", "/login**", "/index.html", "/home.html").permitAll()
            .anyRequest().authenticated()
        .and().exceptionHandling().authenticationEntryPoint(new AuthenticationProcessingFilterEntryPoint("/"))
        .and().logout().logoutSuccessUrl("/").permitAll()
        .and().csrf().csrfTokenRepository(csrfTokenRepository())
        .and().addFilterAfter(csrfHeaderFilter(), CsrfFilter.class)
        .addFilterBefore(ssoFilter(), BasicAuthenticationFilter.class);
        // @formatter:on
    }

    public class AuthenticationProcessingFilterEntryPoint extends LoginUrlAuthenticationEntryPoint {
        public AuthenticationProcessingFilterEntryPoint(String loginFormUrl) {
            super(loginFormUrl);
        }

        @Override
        public void commence(HttpServletRequest request, HttpServletResponse response,
                AuthenticationException authException) throws IOException, ServletException {
            RedirectStrategy redirectStrategy = new DefaultRedirectStrategy();
            redirectStrategy.sendRedirect(request, response, getLoginFormUrl() + "?" + request.getQueryString());
        }
    }

有什么问题吗？

Answer 1

我是这样实现的:

    private Filter ssoFilter(ClientResources client, String path) {
        OAuth2ClientAuthenticationProcessingFilter clientFilter = new OAuth2ClientAuthenticationProcessingFilter(path);
        .......
        clientFilter.setAuthenticationSuccessHandler(new UrlParameterAuthenticationHandler());
        return clientFilter;
    }

    public class UrlParameterAuthenticationHandler extends SimpleUrlAuthenticationSuccessHandler {

        @Override
        protected void handle(HttpServletRequest request, HttpServletResponse response, Authentication authentication)
                throws IOException, ServletException {
            String targetUrl = determineTargetUrl(request, response);

            if (response.isCommitted()) {
                logger.debug("Response has already been committed. Unable to redirect to " + targetUrl);
                return;
            }

            String queryString = HttpUtils.removeParams(request.getQueryString(), "state", "code");
            targetUrl = !StringUtils.isEmpty(queryString) ? targetUrl + "?" + queryString : targetUrl;
            getRedirectStrategy().sendRedirect(request, response, targetUrl);
        }

    }

如果有更好的方法请指正