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>

Ajax请求Session超时解决