tomcat4 中的 web.xml如何配置啊？ 在线等待

Question

<?xml version="1.0" encoding="ISO-8859-1"?><!DOCTYPE web-app
    PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN"
    "http://java.sun.com/j2ee/dtds/web-app_2_2.dtd"><web-app>
<servlet>
<servlet-name>replets</servlet-name>
<servlet-class>inetsoft.sree.web.ServletRepository</servlet-class>
<init-param>
<param-name>sree.home</param-name>
<param-value>C:\StyleReportEELite/Tomcat/webapps/sree/WEB-INF/classes</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>manager</servlet-name>
<servlet-class>inetsoft.sree.adm.AdmServlet</servlet-class>
<init-param>
<param-name>sree.home</param-name>
<param-value>C:\StyleReportEELite/Tomcat/webapps/sree/WEB-INF/classes</param-value>
</init-param>
</servlet>
<servlet-mapping>
   <servlet-name>replets</servlet-name>
   <url-pattern>/replets</url-pattern>
</servlet-mapping>
<servlet-mapping>
   <servlet-name>manager</servlet-name>
   <url-pattern>/manager</url-pattern>
</servlet-mapping>
</web-app>

Answer 1

楼上的，应该把testServlet插入到哪里呢？

Answer 2

几个小时之后，还是自己解决了:现公布如下：
 <servlet> 
       <servlet-name> 
          testServlet 
       </servlet-name> 
       <servlet-class> 
          testServlet 
       </servlet-class> 
     </servlet>
     <servlet-mapping> 
        <servlet-name> 
           testServlet 
        </servlet-name> 
        <url-pattern> 
         testServlet 
        </url-pattern> 
     </servlet-mapping> 参考网址：http://www.ccw.com.cn/htm/center/prog/02_8_1_3.asp
 相信对大家有用

Answer 3

<servlet> 
       <servlet-name>  
          testServlet  Servlet 的名称
       </servlet-name> 
       <servlet-class> 
          testServlet  Servlet  的类名称
       </servlet-class> 
     </servlet>
     <servlet-mapping> 
        <servlet-name> 
           testServlet Servlet 映射的类名称 
        </servlet-name> 
        <url-pattern> 
         /testServlet  
        </url-pattern> Servlet 映射的虚拟路径名称 
     </servlet-mapping>

Answer 4

<servlet>
        <servlet-name>testServlet</servlet-name>
        <servlet-class>testServlet1</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>testServlet</servlet-name>
        <url-pattern>/testServlet</url-pattern>
    </servlet-mapping>
真正正确的设法,这种设法实际上石油入口参数的Servlet,对于没有参数的Servlet
没有必要去改动web.xml