在 Spring MVC 中检测 Ajax 请求中的会话超时

Question

我似乎找不到关于如何在会话超时时从 ajax 请求发回一些数据的好示例/答案。它发送回登录页面 HTML，我想发送 json 或我可以拦截的状态代码。

Answer 1

最简单的方法是对 AJAX 请求的 URL 使用过滤器。

在下面的示例中，我只是发送 HTTP 500 响应代码，并带有指示会话超时的响应正文，但您可以轻松地将响应代码和正文设置为更适合您的情况。

package com.myapp.security.authentication;

import org.springframework.web.filter.GenericFilterBean;

import javax.servlet.FilterChain;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;

public class ExpiredSessionFilter extends GenericFilterBean {

    static final String FILTER_APPLIED = "__spring_security_expired_session_filter_applied";

    @Override
    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException {

        HttpServletRequest request = (HttpServletRequest) req;
        HttpServletResponse response = (HttpServletResponse) res;

        if (request.getAttribute(FILTER_APPLIED) != null) {
            chain.doFilter(request, response);
            return;
        }

        request.setAttribute(FILTER_APPLIED, Boolean.TRUE);
        if (request.getRequestedSessionId() != null && !request.isRequestedSessionIdValid()) {               
            response.sendError(HttpServletResponse.SC_INTERNAL_SERVER_ERROR, "SESSION_TIMED_OUT");
            return;
        }

        chain.doFilter(request, response);
    }
}

Answer 2

这是我认为非常简单的方法。这是我在该网站上观察到的方法的组合。我写了一篇关于它的博客文章：
http://yoyar. com/blog/2012/06/dealing-with-the-spring-security-ajax-session-timeout-problem/

基本思想是使用上面建议的 api url 前缀（即 /api/secured）以及身份验证入口点。它简单且有效。

这是身份验证入口点：

package com.yoyar.yaya.config;

import org.springframework.security.core.AuthenticationException;
import org.springframework.security.web.authentication.LoginUrlAuthenticationEntryPoint;

import javax.servlet.ServletException;
import javax.servlet.http.*;
import java.io.IOException;

public class AjaxAwareAuthenticationEntryPoint 
             extends LoginUrlAuthenticationEntryPoint {

    public AjaxAwareAuthenticationEntryPoint(String loginUrl) {
        super(loginUrl);
    }

    @Override
    public void commence(
        HttpServletRequest request, 
        HttpServletResponse response, 
        AuthenticationException authException) 
            throws IOException, ServletException {

        boolean isAjax 
            = request.getRequestURI().startsWith("/api/secured");

        if (isAjax) {
            response.sendError(403, "Forbidden");
        } else {
            super.commence(request, response, authException);
        }
    }
}

这是 spring 上下文 xml 中的内容：

<bean id="authenticationEntryPoint"
  class="com.yoyar.yaya.config.AjaxAwareAuthenticationEntryPoint">
    <constructor-arg name="loginUrl" value="/login"/>
</bean>

<security:http auto-config="true"
  use-expressions="true"
  entry-point-ref="authenticationEntryPoint">
    <security:intercept-url pattern="/api/secured/**" access="hasRole('ROLE_USER')"/>
    <security:intercept-url pattern="/login" access="permitAll"/>
    <security:intercept-url pattern="/logout" access="permitAll"/>
    <security:intercept-url pattern="/denied" access="hasRole('ROLE_USER')"/>
    <security:intercept-url pattern="/" access="permitAll"/>
    <security:form-login login-page="/login"
                         authentication-failure-url="/loginfailed"
                         default-target-url="/login/success"/>
    <security:access-denied-handler error-page="/denied"/>
    <security:logout invalidate-session="true"
                     logout-success-url="/logout/success"
                     logout-url="/logout"/>
</security:http>

Answer 3

我在后端使用@Matt 的相同解决方案。如果您在前端使用 AngularJs，请在 Angular $http 中添加以下拦截器，让浏览器实际重定向到登录页面。

var HttpInterceptorModule = angular.module('httpInterceptor', [])
.config(function ($httpProvider) {
  $httpProvider.interceptors.push('myInterceptor');
  $httpProvider.defaults.headers.common["X-Requested-With"] = 'XMLHttpRequest'; 
})
 .factory('myInterceptor', function ($q) {
return {
    'responseError': function(rejection) {
      // do something on error
        if(rejection.status == 403 || rejection.status == 401) window.location = "login";   
        return $q.reject(rejection);
    }
  };

});

请注意，仅当您在 1.1.1 版本之后使用 AngularJs 时才需要下面的行（AngularJS 从该版本开始删除了标头“X-Requested-With”）

$httpProvider.defaults.headers.common["X-Requested-With"] = 'XMLHttpRequest';

Answer 4

鉴于目前所有的答案都已经有几年了，我将分享我目前在 Spring Boot REST 应用程序中工作的解决方案：

@Configuration
@EnableWebSecurity
public class UISecurityConfig extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        ...
        http.exceptionHandling.authenticationEntryPoint(authenticationEntryPoint());
        ...
    }

    private AuthenticationEntryPoint authenticationEntryPoint() {
        // As a REST service there is no 'authentication entry point' like MVC which can redirect to a login page
        // Instead just reply with 401 - Unauthorized
        return (request, response, authException) -> response.sendError(HttpServletResponse.SC_UNAUTHORIZED, authException.getMessage());
    }
}

这里的基本前提是我覆盖默认情况下发出的身份验证入口点重定向到我不存在的登录页面。现在，它通过发送 401 进行响应。Spring 还隐式创建一个它也返回的标准错误响应 JSON 对象。