web service服务端如何编写web.xml - 编程论坛

#2

lampeter1232011-01-19 17:07

用MyEclipse建立项目就能自动生成web.xml

程序代码：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java. xmlns:xsi="http://www. version="2.4" xsi:schemaLocation="http://java. <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>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>

#3

wzh5453946222011-01-20 00:28

程序代码：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java. xmlns:xsi="http://www. version="2.5" xsi:schemaLocation="http://java. <servlet-name>XFireServlet</servlet-name>
    <servlet-class>org.codehaus.xfire.transport.http.XFireConfigurableServlet</servlet-class>
    <load-on-startup>0</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>XFireServlet</servlet-name>
    <url-pattern>/services/*</url-pattern>
  </servlet-mapping>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>

#4

w524804042011-04-21 22:25

楼上二位发的我看的不是很理解能解释解释么？