利用URLrewrite实现伪静态地址输出

今天项目过程当中发现不少超连接把参数所有暴露在浏览器地址栏，甚是担忧。幸亏，被我发现了一种实现技术，哈哈。。。。

1.下载urlrewrite,官方下载地址：http://tuckey.org/urlrewrite/dist/urlrewritefilter-2.6.zip  有可能直接下载不了，你能够找个×××代理网站打开下载。

如：http://www.cnproxy.com/webproxy.php 

下载获得urlrewritefilter-3.2.0.zip 解压后 找到lib下的urlrewrite-3.2.0.jar包拷贝到你的工程lib下，拷贝urlrewrite.xml到web_INF下。

2.在web.xml中Filter的形式加载urlrewrite注明过滤的内容:

<filter>
        <filter-name>UrlRewriteFilter</filter-name>
        <filter-class>
            org.tuckey.web.filters.urlrewrite.UrlRewriteFilter
        </filter-class>
    </filter>
    <filter-mapping>
        <filter-name>UrlRewriteFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

3.修改urlrewrite.xml

<?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
http://tuckey.org/urlrewrite/

-->
<urlrewrite>
    <rule>
        <from>^/([a-z]+)$</from>
        <to type= "forward" >/world.jsp?id=$1</to>
    </rule>
    <rule>
        <from>^/world/(.*)$</from>
        <to>/world.jsp?tid=$1</to>
    </rule>
    <rule>
        <from>^/(.*).html$</from>
        <to>/test1/$1.jsp</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>
</urlrewrite>

rule是url重写规则,from是显示出来的地址,to是映射的实际地址,$1是重写参数,能够为多个,()里是匹配的正则表达式.
好了,在项目中新建world.jsp,启动tomcat,输入
http://localhost:8080/Uset/world/1

Uset是你的工程名

实际上访问的是http://localhost:8080/Uset/world.jsp?tid=1
这样就简单的实现了伪静态的效果