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JSP页面URL重写(URL Rewrite)
java技术 2008-09-20 18:37:57 阅读60 评论1 字号：大中小
这段时间一直都在研究jsp生成静态页面的方法，在网上找了很多资料，也有不少朋友给我提出解决方案，确实解决了不少的问题，但是这样做有点麻烦。其实我并不是想减轻服务器的压力，我们要做的不是门户网站，访问量没那么大，所以不用担心服务器的承受能力。只是希望搜索引擎能够搜索到我们的页面，只要能达到目的，用什么样的方式都可以。
在网上看到了一则URL重写的贴子，感觉比较适合我这样的情况，应用起来简单。然后自己试着写了一个例子，居然成功了，而在Struts里不知道怎么实现，实现转发的时候好象没什么用了。*.do好象不能实现映射
以下是相关代码，若有不正之处，欢迎大家指正!
首先要去下载一个urlrewritefilter-2.6.zip,然后把它解压到WEB-INF下，然后配置一下web.xml
<?xml version=”1.0″ encoding=”UTF-8″?>
<web-app xmlns=”http://java.sun.com/xml/ns/j2ee” xmlns:xsi=”http://www.w3.org/2001/XMLSchema-instance” version=”2.4″ xsi:schemaLocation=”http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd“>
<filter>
<filter-name>UrlRewriteFilter</filter-name>
<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
<init-param>
<param-name>logLevel</param-name>
<param-value>WARN</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>UrlRewriteFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>action</servlet-name>
<servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
<init-param>
<param-name>config</param-name>
<param-value>/WEB-INF/struts-config.xml</param-value>
</init-param>
<init-param>
<param-name>debug</param-name>
<param-value>3</param-value>
</init-param>
<init-param>
<param-name>detail</param-name>
<param-value>3</param-value>
</init-param>
<load-on-startup>0</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>action</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
然后再随便建立几个jsp页面，如：
MyJsp.jsp
<%@ page language=”java” import=”java.util.*” pageEncoding=”GBK”%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+”://”+request.getServerName()+”:”+request.getServerPort()+path+”/”;
%>
<!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01 Transitional//EN”>
<html>
<head>
<base href=”<%=basePath%>”>
<title>My JSP ‘MyJsp.jsp’ starting page</title>
<meta http-equiv=”pragma” content=”no-cache”>
<meta http-equiv=”cache-control” content=”no-cache”>
<meta http-equiv=”expires” content=”0″>
<meta http-equiv=”keywords” content=”keyword1,keyword2,keyword3″>
<meta http-equiv=”description” content=”This is my page”>
<!–
<link rel=”stylesheet” type=”text/css” href=”styles.css”>
–>
</head>
<body>
<%
String a = request.getParameter(”id”);
%>
<%=basePath %> <br>
<%
if(a.equals(”123″))
{
out.println(”哈哈”);
}
else
{
out.println(”再试一次!”);
}
%>
</body>
</html>
接着在urlrewrite.xml里配置一下路径
<?xml version=”1.0″ encoding=”utf-8″?>
<!DOCTYPE urlrewrite PUBLIC “-//tuckey.org//DTD UrlRewrite 2.6//EN”
“http://tuckey.org/res/dtds/urlrewrite2.6.dtd“>
<!–
Configuration file for UrlRewriteFilter
–>
<urlrewrite>
<rule>
<note>
The rule means that requests to /test/status/ will be redirected to /rewrite-status
the url will be rewritten.
</note>
<from>/test/status/</from>
<to type=”redirect”>%{context-path}/rewrite-status</to>
</rule>
<outbound-rule>
<note>
The outbound-rule specifies that when response.encodeURL is called (if you are using JSTL c:url)
the url /rewrite-status will be rewritten to /test/status/.
The above rule and this outbound-rule means that end users should never see the
url /rewrite-status only /test/status/ both in thier location bar and in hyperlinks
in your pages.
</note>
<from>/rewrite-status</from>
<to>/test/status/</to>
</outbound-rule>
<rule>
<from>/test/([0-9]+)</from>
<to>/MyJsp.jsp?id=$1</to>
</rule>
</urlrewrite>
启动服务器，然后输入http://localhost:8080/test/123 ,那么就可以显示了,而实际上读取的路径是http://localhost:8080/MyJsp.jsp?id=123
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