启动tomcat时，怎样启动一个servlet！急！在线等

Question

可以参考下面这个格式:
<servlet>
    <servlet-name>JPEGServlet</servlet-name>
    <servlet-class>com.jspdev.ch7.JPEGServlet</servlet-class>
    </servlet>
     <servlet>
        <servlet-name>JDBCServlet</servlet-name>
        <servlet-class>com.jspdev.ch7.JDBCServlet</servlet-class>
         <init-param>
         <param-name>DRIVER</param-name>
         <param-value>com.microsoft.jdbc.sqlserver.SQLServerDriver</param-value>
        </init-param>
        <init-param>
         <param-name>URL</param-name>
         <param-value>jdbc:microsoft:sqlserver://127.0.0.1:1433;DatabaseName=jspdev</param-value>
        </init-param>
         <init-param>
         <param-name>USER</param-name>
         <param-value>bn</param-value>
        </init-param>
         <init-param>
         <param-name>PASSWORD</param-name>
         <param-value>bn</param-value>
        </init-param>
    </servlet>

Answer 1

<servlet>
<servlet-name>OnLineCounterLisenter</servlet-name>
    <display-name>OnLineCounterLisenter</display-name>
    <servlet-class>web80info.framework.util.OnLineCounterLisenter</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>给个例子，
这个类在服务器启动的时候就生成一个实例．一般可以在里面初始化一些参数等等