java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/SpringDispatcher-ser


When I launch my application I get this error

java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/SpringDispatcher-servlet.xml] nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/SpringDispatcher-servlet.xml]

Meanwhile, I dont have any file like SpringDispatcher-servlet.xml neither do I in my web.xml or mvc-dispatcher-servlet.xml file defined in my WEB-INF folder.

<strong>web.xml file</strong>

<context-param> <param-name>contextClass</param-name> <param-value> org.springframework.web.context.support.AnnotationConfigWebApplicationContext </param-value> </context-param> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <servlet> <servlet-name>SpringDispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>SpringDispatcher</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value> </context-param> <session-config> <session-timeout>30</session-timeout> </session-config>

<strong>mvc-dispatcher-servlet.xml file</strong>

<context:component-scan base-package="aish.vaishno.musicstore.controller" /> <mvc:annotation-driven /> <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="prefix" value="/WEB-INF/view/" /> <property name="suffix" value=".jsp" /> </bean>

Please how can I locate this file


Please what am I getting wrong?


Spring is looking SpringDispatcher-servlet.xml in your web project and as it is unable to find it, it is throwing an exception.

You can override the dispatcher servlet xml file like this - providing blank arguments.

<servlet> <servlet-name>SpringDispatcher</servlet-name> <servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value></param-value> </init-param> </servlet>


When you define Dispatcher servlet in web.xml spring expects Web application context name as Disptacherservletname-servlet.xml under /WEB-INF/ . In your case it should be SpringDispatcher-servlet.xml not mvc-dispatcher-servlet.xml

Or you can use contextConfigLocation parameter to follow your own naming conventions.

<servlet> <servlet-name>SpringDispatcher</servlet-name> <servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value> </init-param> </servlet>


In your web.xml you have define the servlet to be

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

therefore, you nee to create a file named <strong>SpringDispatcher-servlet.xml</strong> inside your <strong>WEB-INF</strong> folder. This is just how it works. If you change the servlet-name to be dispatcher then the file name should be dispatcher-servlet.xml. Your SpringDispatcher-servlet.xml contains the definitions of your spring context. Take a look on this <a href="https://javapointers.com/tutorial/create-spring-mvc-project-using-maven/" rel="nofollow">tutorial</a>.


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