使用场景:java
在现实项目中,偶尔须要在 跳转 指定的 项目首页以前作一些处理,可是又不想 单独多加一个 跳转页面来完成这次操做;如此,该如何处理呢?web
这里,能够 利用 web.xml 中的 welcome-file-list 设置默认 来 处理。具体操做以下面的例子所示:app
注意: welcome-file 节点 须要 与 servlet-mapping 中的 url-pattern 节点 内容保持一致.this
一、web.xml url
<?xml version="1.0" encoding="UTF-8"?> <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> <display-name>web_test</display-name> <welcome-file-list> <welcome-file>emp_tz.hts</welcome-file> </welcome-file-list> <servlet> <servlet-name>FrameControlServlet</servlet-name> <servlet-class>com.myweb.test.MyDefaultManagerServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>FrameControlServlet</servlet-name> <url-pattern>/emp_tz.hts</url-pattern> </servlet-mapping> </web-app>
二、对应 servlet 处理spa
package com.myweb.test; import java.io.IOException; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; public class MyDefaultManagerServlet extends HttpServlet{ private static final long serialVersionUID = 1L; protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { System.out.println("this is doPost..."); response.getWriter().write("very ok"); } protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { System.out.println("this is doGet..."); response.getWriter().write("ok"); } }