Filter简单过滤器的实现

Posted 2020-06-21 太快的幸福,措手不及

1.过滤器简单编写

package cn.buaa.core.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;

import cn.buaa.core.constant.Constant;

public class LoginFilter implements Filter {

    @Override
    public void destroy() {
        
    }

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain chain) throws IOException, ServletException {
        HttpServletRequest request = (HttpServletRequest) servletRequest;
        HttpServletResponse response = (HttpServletResponse)servletResponse;
        String uri = request.getRequestURI();
        //判断当前请求地址是否是登录的请求地址
        if(!uri.contains("sys/login_")){
            //非登录请求
            if(request.getSession().getAttribute(Constant.USER) != null){
                //说明已经登陆过，放行
                chain.doFilter(request, response);
            }else{
                //没有登录，跳转到登录页面
                response.sendRedirect(request.getContextPath() + "/sys/login_toLoginUI.action");
            }
        }else{
            //登录请求，直接放行
            chain.doFilter(request, response);
        }
    }

    @Override
    public void init(FilterConfig arg0) throws ServletException {
        
    }

}

2.xml中进行配置（注意：必须配置在struts的前边，要不然不起作用）

<filter>
        <filter-name>loginFilter</filter-name>
        <filter-class>cn.buaa.core.filter.LoginFilter</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>loginFilter</filter-name>
        <url-pattern>*.action</url-pattern>
    </filter-mapping>