jsp 用urlrewrite 实现URL 重写是否看到别人的网站网址名都不带后缀名比较酷,比如qq空间的地址,其实用urlrewrite这个包很容易就实现了。下面是使用说明: 1.下载urlrewrite,官方下载地址:http://tuckey.org/urlrewrite/dist/urlrewritefilter-2.6.zip 2.解压缩文件,压缩包内文件copy到项目中(压缩包位置 -> 项目位置): urlrewrite-2.6.0-src/webapp/WEB-INF/lib/urlrewrite-2.6.0.jar -> WebRoot/WEB-INF/lib/urlrewrite-2.6.0.jar urlrewrite-2.6.0-src/webapp/WEB-INF/urlrewrite.xml -> WebRoot/WEB-INF/urlrewrite.xml 3.将以下代码添加到web.xml里 Xml代码
<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> 4.修改urlrewrite.xml Xml代码
<?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,可以为多个,()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/mysite/world/1 mysite是你的项目名 实际上访问的是http://localhost:8080/mysite/world.jsp?tid=1
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