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

没有正确配置applicationContext.xml位置引起的。
所以要自己增加参数

<init-param>
			<param-name>contextConfigLocation</param-name>
			<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
		</init-param>

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
	http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
	<display-name></display-name>


	<servlet>  
        <servlet-name>spmvc</servlet-name>  
        <servlet-class>  
            org.springframework.web.servlet.DispatcherServlet  
        </servlet-class>  
        <init-param>
			<param-name>contextConfigLocation</param-name>
			<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
		</init-param>
        <load-on-startup>1</load-on-startup>  
    </servlet>  
    <servlet-mapping>  
        <servlet-name>spmvc</servlet-name>  
        <url-pattern>*.do</url-pattern>  
    </servlet-mapping>


	<welcome-file-list>
		<welcome-file>index.jsp</welcome-file>
	</welcome-file-list>
</web-app>

或者在servlet定义的前面加：

<context-param>
    	<param-name>contextConfigLocation</param-name>
		<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
    </context-param>