关于OC4J与servlet的最基本的问题！

Question

web.xml作如下配置,能看懂吧,访问的时候用http://localhost/testservlet
<web-app>
  <servlet>
    <servlet-name>testservlet</servlet-name>
    <servlet-class>testwar.testServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>testservlet</servlet-name>
    <url-pattern>/testservlet</url-pattern>
  </servlet-mapping>
</web-app>

Answer 1

我试过了，做了类似的改动后
<web-app>
<!-- A demo servlet, add servlets below -->
<servlet>
<servlet-name>receiveAnswer</servlet-name>
<servlet-class>receiveAnswer</servlet-class>
</servlet>
<servlet>
<servlet-name>snoop</servlet-name>
<servlet-class>SnoopServlet</servlet-class>
</servlet>
</web-app>
重启服务的时候系统提示错误

Answer 2

类放到classes目录下了吗?
<servlet-class>要写完整路径,包括包名

Answer 3

我把它放到了classes目录下但是我的程序里没有用到包
这样改对吗？

Answer 4

对,完整的web.xml文件
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<!-- A demo servlet, add servlets below -->
<servlet>
<servlet-name>receiveAnswer</servlet-name>
<servlet-class>receiveAnswer</servlet-class>
</servlet>
<servlet>
<servlet-name>snoop</servlet-name>
<servlet-class>SnoopServlet</servlet-class>
</servlet>
</web-app>
再不行仔细检查名称,包括大小写

Answer 5

我的程序
<?xml version="1.0"?>
<!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>
<!-- A demo servlet, add servlets below -->
<servlet>
<servlet-name>receiveAnswer</servlet-name>
<servlet-class>receiveAnswer</servlet-class>
</servlet>
<servlet>
<servlet-name>snoop</servlet-name>
<servlet-class>SnoopServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>receiveAnswer</servlet-name>
    <url-pattern>/receiveAnswer</url-pattern>
  </servlet-mapping>
</web-app>
路径http://000.000.00.00/receiveAnswer
错误提示：因为地址不正确，无法显示您要查看的页。
帮忙呀!我快死了

Answer 6

<servlet>
<servlet-name>receiveAnswer</servlet-name>
<servlet-class>receiveAnswer</servlet-class>
</servlet>
<servlet>
<servlet-name>snoop</servlet-name>
<servlet-class>SnoopServlet</servlet-class>
</servlet>

Answer 7

好了我的问题解决了,web.xml没有问题！
谢谢大家的帮助，谢谢！！！