Question

Servlet中的web.xml文件怎么配置

Answer 1

不知道是不是你想要的

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="扰肆2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<servlet>
<servlet-name>action</servlet-name>消李羡
<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>拿拍0</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>action</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>