网站缓存数据到tomcat服务器

来源：互联网发布：苹果6怎么开4g网络编辑：程序博客网时间：2024/06/05 22:52

通过缓存使相同的数据不用重复加载，降低数据库的访问

public class CacheFilter implements Filter {  //实例变量【每线程共享】private Map<String,byte[]> cache=new HashMap<String , byte[]>();@Overridepublic void destroy() {}@Overridepublic void doFilter(ServletRequest req, ServletResponse res,FilterChain chain) throws IOException, ServletException {HttpServletRequest request=(HttpServletRequest) req;HttpServletResponse response=(HttpServletResponse) res;//1、取得客户端访问路径String uri=request.getRequestURI();//2、根据URL去缓存中查询对应的页面资源byte[] data=cache.get(uri);//3、如果没有就放行到web资源中if(data==null){MyResponse myResponse=new MyResponse(response);chain.doFilter(request, response);//4、返回后将缓存数据放到缓存中，便于下次使用data=myResponse.getBuffer();//6、向浏览器输出cache.put(uri, data);System.out.println("从服务端获取资源");}else{//5、如果有，就直接从缓存中取得数据System.out.println("从缓存读取");chain.doFilter(request, response);}}@Overridepublic void init(FilterConfig arg0) throws ServletException {}}

import java.io.IOException;import java.io.PrintWriter;import javax.servlet.ServletException;import javax.servlet.ServletOutputStream;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;public class ShowServlet extends HttpServlet {public void doGet(HttpServletRequest request, HttpServletResponse response)throws ServletException, IOException {ServletOutputStream sout=response.getOutputStream();sout.write("fgfgdfsgdsfgdsfgdsfgdsfgsdfgdsdsfgdsfgsdfgdsfgsdfgkjslkdfjgl;kdsjglsdjfglsdjfgldfsjglksdfjgsfdl".getBytes());}public void doPost(HttpServletRequest request, HttpServletResponse response)throws ServletException, IOException {}}

0 0