Ajax请求session超时解决web前端js代码：
$.ajaxSetup({ contentType : "application/x-www-form-urlencoded;charset=utf-8", complete : function(xhr, textStatus) { if (xhr.status == 520) {//如果返回状态码是520 window.location..reload();//刷新页面，执行登录逻辑 return; } }});
java代码：
1. 写一个filter
import java.io.IOException;import javax.servlet.Filter;import javax.servlet.FilterChain;import javax.servlet.FilterConfig;import javax.servlet.ServletException;import javax.servlet.ServletRequest;import javax.servlet.ServletResponse;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;public class SessionTimeoutFilter implements Filter { public void destroy() { } public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException { HttpServletRequest req = (HttpServletRequest) request; HttpServletResponse res = (HttpServletResponse) response; // 判断session里是否有用户信息 if (req.getSession().getAttribute("username") == null){ // 如果是ajax请求响应头会有，x-requested-with； if (req.getHeader("x-requested-with") != null && req.getHeader("x-requested-with").equalsIgnoreCase("xmlHttPRequest")){ res.setStatus(520);//表示session timeout }else{ chain.doFilter(req, res); } }else{ chain.doFilter(req, res); } } public void init(FilterConfig chain) throws ServletException { }}
2. 在web.xml中添加上面的filter
<filter> <filter-name>ajaxSessionTimeout</filter-name> <filter-class>org.tshark.framework.web.filter.SessionTimeoutFilter</filter-class></filter><filter-mapping> <filter-name>ajaxSessionTimeout</filter-name> <url-pattern>/*</url-pattern></filter-mapping>