Question

java web开发中怎样在web.xml文件中配置一个普通的servlet<*.do> 要求：此servlet要拦截所有XXX.do的请求，

Answer 1

<servlet>
<servlet-name>servletName</启余servlet-name>
<servlet-class>com.xxx.xxx.servletClass</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>servletName</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>

web.xml里面的过滤器配置只支持以下三种
①　完全匹配
<url-pattern>/test/zhidao.do</url-pattern>
② 目录匹配
<url-pattern>/test/*</url-pattern>
③ 扩展名匹配
<url-pattern>伍旁陪*.do</url-pattern>

而不支持腔蠢/*.do这样的

Answer 2

<servlet>
<servlet-name>servlet名称</servlet-name>卜磨
<servlet-class>知弊型包.类名</servlet-class>搭猜
</servlet>

<servlet-mapping>
<servlet-name>servlet名称</servlet-name>
<url-pattern>/*.do</url-pattern>
</servlet-mapping>

Answer 3

<servlet>
<servlet-name>servletName</servlet-name>
<servlet-class>com.xxx.xxx.servletClass</servlet-class>
</servlet>键缺芦

<servlet-mapping>
<servlet-name>扮宽servletName</servlet-name>
<url-pattern>/*.do</稿带url-pattern>
</servlet-mapping>