web.xml问题!!!
我用webwork+spirng+hibernate写的一些程序,现在想和以前用spring+hibernate
整合.但是现在怎么启动不了.好像是web.xml问题!
我的web.xml

<?xml version= "1.0 " encoding= "UTF-8 "?>
<web-app version= "2.4 "
xmlns= "http://java.sun.com/xml/ns/j2ee "
xmlns:xsi= "http://www.w3.org/2001/XMLSchema-instance "
xsi:schemaLocation= "http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd ">
<!-- 定义分发器 -->
<filter>
<filter-name> webwork </filter-name>
<filter-class>
com.opensymphony.webwork.dispatcher.FilterDispatcher
</filter-class>
</filter>

<!-- 定义过滤器,禁止非登陆直URL进入 -->
<!-- <filter>
<filter-name> securityFilter </filter-name>
<filter-class> com.sztelecom.csc.mcms.tools.SecurityFilter </filter-class>
</filter>
-->
<context-param>
<param-name> contextConfigLocation </param-name>
<param-value> /WEB-INF/classes/applicationContext.xml </param-value>
</context-param>

<!-- 分发器拦截映射 -->
<filter-mapping>
<filter-name> webwork </filter-name>
<url-pattern> /* </url-pattern>
</filter-mapping>
<!-- 拦截器映射
<filter-mapping>
<filter-name> securityFilter </filter-name>
<url-pattern> /jsp/* </url-pattern>
</filter-mapping>
-->

<listener>
<listener-class> org.springframework.web.context.ContextLoaderListener </listener-class>
</listener>

<!-- aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -->

<context-param>
<param-name> contextConfigLocation </param-name>
<param-value> /WEB-INF/conf/dispatcherServletCustomer-servlet.xml </param-value>

</context-param>

<servlet>
<servlet-name> dispatcherServlet </servlet-name>
<servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class>

<load-on-startup> 1 </load-on-startup>
</servlet>



<servlet-mapping>
<servlet-name> dispatcherServlet </servlet-name>
<url-pattern> *.do </url-pattern>
</servlet-mapping>

<listener>
<listener-class> org.springframework.web.context.ContextLoaderListener </listener-class>
</listener>



<servlet>
<servlet-name> webwork </servlet-name>
<servlet-class> com.opensymphony.webwork.dispatcher.ServletDispatcher </servlet-class>
</servlet>
<servlet-mapping>
<servlet-name> webwork </servlet-name>

<url-pattern> *.action </url-pattern>
</servlet-mapping>
<servlet>
<servlet-name> context </servlet-name>
<servlet-class> org.springframework.web.context.ContextLoaderServlet </servlet-class>
<load-on-startup> 2 </load-on-startup>
</servlet>

</web-app>

aaaaa的上面是webwork+spring+hibnernate写的web.xml
aaaaa的下面是用spring+hibernate写的..
求教!!!

[解决办法]
Caused by: java.lang.IllegalArgumentException: Duplicate context initialization parameter contextConfigLocation
你写了2格一样的参数contextConfigLocation