Could not open ServletContext resource [/WEB-INF/dispatcherservlet-servlet.xml]

今天在研究Spring mvc 的时候碰到了这样的一个问题：
Could not open ServletContext resource [/WEB-INF/dispatcherservlet-servlet.xml]

原因是我的web.xml里面是这样配置的：

[html]  view plain  copy

1. <servlet>  
2.     <servlet-name>dispatcherservlet</servlet-name>  
3.     <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  
4.     <init-param>  
5.         <param-name>springmvcconfig</param-name>  
6.         <param-value>/WEB-INF/springmvc-config.xml</param-value>  
7.     </init-param>  
8.     <load-on-startup>1</load-on-startup>  
9. </servlet>  
10. <servlet-mapping>  
11.     <servlet-name>dispatcherservlet</servlet-name>  
12.     <url-pattern>*.do</url-pattern>  
13. </servlet-mapping>  

这样配置的话，默认情况下回去读取dispatcherservlet-servlet.xml，所有改改配置才对。

[html]  view plain  copy

1. <servlet>  
2.     <servlet-name>dispatcherservlet</servlet-name>  
3.     <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  
4.     <init-param>  
5.         <param-name>contextConfigLocation</param-name>  
6.         <param-value>/WEB-INF/springmvc-config.xml</param-value>  
7.     </init-param>  
8.     <load-on-startup>1</load-on-startup>  
9. </servlet>  
10. <servlet-mapping>  
11.     <servlet-name>dispatcherservlet</servlet-name>  
12.     <url-pattern>*.do</url-pattern>  
13. </servlet-mapping>  

即：把

[html]  view plain  copy

1. <param-name>contextConfigLocation</param-name>  

这里配置成这样才能读取到你配置的spring mvc 的配置

转载：http://blog.csdn.net/limingchuan123456789/article/details/8886095

参考：http://www.oschina.net/question/158170_38219