Context initialization failed in spring4

Question

I am getting the following error when running my project in maven.

org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/dispatcher-servlet.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/dispatcher-servlet.xml]

I have tried the solution specified on the internet like servlet name should be same as the file in which we will specify context param but still it is not working.

Web.xml

<web-app>
  <display-name>Archetype Created Web Application</display-name>
  <!-- <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/dispatcher.xml</param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener> -->
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <!-- <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/dispatcher-servlet/dispatcher-servlet.xml</param-value>
        </init-param> -->
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

dispatcher.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->
    <annotation-driven />

    <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <resources mapping="/resources/**" location="/resources/" />

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/" />
        <beans:p`enter code here`roperty name="suffix" value=".jsp" />
    </beans:bean>
    <context:component-scan base-package="com.test.controller" />
</beans:beans>

Answer 1

put your dispatcher-servlet.xml file directly inside the WEB-INF directory where the web.xml is placed. (I can provide a detailed answer if you need)

    <web-app>
  <display-name>Archetype Created Web Application</display-name>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener> -->
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>

        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>