tomcat服务器下的webapps目录下布置应用？？？找不到页面！请教！（内有详细说明）

[pipilu1978]

我在webapps目录下布置了一个目录Family_Center目录结构如下所示
Family_Center-->
html-->
Entrance.html
images-->
*.jpg
jsp-->
app-->
*.jsp
WEB-INF-->
classes-->(类包）
web.xml
然后我在conf目录下的server.xml中加了这样一句
<Context path="/Family_Center" docBase="webapps/Family_Center" crossContext="true" debug="0"
reloadable="true" trusted="false">
</Context>
另外把web.xml代码也贴出来，原因是要用到servlet做控制器，具体部署我不太了解！xml代码看不懂！请教高手详细讲解！我在ie中输入http://localhost:8080/Family_Center,报错找不到Family_Center，请高手帮忙指点
<?xml version="1.0" encoding="UTF8" ?>
<!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>

<!-- Welcome file -->
<welcome-file-list>
<welcome-file>/html/Entrance.html</welcome-file>
</welcome-file-list>

<!-- Dispatcher servlet and its init parameters -->
<servlet>
<servlet-name>Dispatcher</servlet-name>
<servlet-class>com.jspbg.servlet.DispatcherServlet</servlet-class>

<init-param>
<param-name>ACTIONS_XML</param-name>
<param-value>/WEB-INF/actions.xml</param-value>
</init-param>
<init-param>
<param-name>AddressBookManager</param-name>
<param-value>com.jspbg.model.AddressBookManagerImpl</param-value>
</init-param>
</servlet>

<!-- Dispatcher servlet mapping -->
<servlet-mapping>
<servlet-name>Dispatcher</servlet-name>
<url-pattern>/Dispatcher</url-pattern>
</servlet-mapping>

<!-- Error jsp -->
<servlet>
<servlet-name>AddressBook</servlet-name>
<jsp-file>/jsp/app/AddressBook.jsp</jsp-file>
</servlet>

<!-- Error jsp URL mapping -->
<servlet-mapping>
<servlet-name>AddressBook</servlet-name>
<url-pattern>/AddressBook</url-pattern>
</servlet-mapping>

<!-- Error jsp -->
<servlet>
<servlet-name>Error</servlet-name>
<jsp-file>/jsp/app/Error.jsp</jsp-file>
</servlet>

<!-- Error jsp URL mapping -->
<servlet-mapping>
<servlet-name>Error</servlet-name>
<url-pattern>/Error</url-pattern>
</servlet-mapping>

<!-- Zero results jsp -->
<servlet>
<servlet-name>ZeroResults</servlet-name>
<jsp-file>/jsp/app/ZeroResults.jsp</jsp-file>
</servlet>

<!-- Zero results jsp URL mapping -->
<servlet-mapping>
<servlet-name>ZeroResults</servlet-name>
<url-pattern>/ZeroResults</url-pattern>
</servlet-mapping>

<!-- Address book results JSP and URL mapping -->
<servlet>
<servlet-name>AddressBookResults</servlet-name>
<jsp-file>/jsp/app/AddressBookResults.jsp</jsp-file>
</servlet>
<servlet-mapping>
<servlet-name>AddressBookResults</servlet-name>
<url-pattern>/AddressBookResults</url-pattern>
</servlet-mapping>

</web-app>

以上代码均出自书上！我在研读代码！

[idilent]

<Context path="/Family_Center" docBase="webapps/Family_Center" crossContext="true" debug="0"
reloadable="true" trusted="false">

改称
<Context path="/Family_Center" docBase="Family_Center" crossContext="true" debug="0"
reloadable="true" trusted="false">

[straybaby]

你没有在web.xml中声明servlet Family_Center.
怎么能用http://localhost:8080/Family_Center访问.
应该有一项
<servlet>
<servlet-name>Family_Center</servlet-name>
<servlet-class>你的servlet的class位置</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Family_Center</servlet-name>
<url-pattern>/Family_Center</url-pattern>
</servlet-mapping>

[pipilu1978]

这个是不是定义Family_Center servlet？to straybaby
<!-- Dispatcher servlet and its init parameters -->
<servlet>
<servlet-name>Dispatcher</servlet-name>
<servlet-class>com.jspbg.servlet.DispatcherServlet</servlet-class>

<init-param>
<param-name>ACTIONS_XML</param-name>
<param-value>/WEB-INF/actions.xml</param-value>
</init-param>
<init-param>
<param-name>AddressBookManager</param-name>
<param-value>com.jspbg.model.AddressBookManagerImpl</param-value>
</init-param>
</servlet>

<!-- Dispatcher servlet mapping -->
<servlet-mapping>
<servlet-name>Dispatcher</servlet-name>
<url-pattern>/Dispatcher</url-pattern>
</servlet-mapping>