I'm quite new to spring and I got stuck just in the beginning... I wanted to create simple app, that would use css and js. But I can't reach that static content... I've found some solutions, but I wasn't able to modify them for my solution
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Would be anybody please so kind and could show some explicit solution (where to add/modify something and what exactly to add/modify) ?
my app structure in navigator:
![Posted Image]()
Content of web.xml
Content of servlet-context.xml
I'm running my app on tomcat server.
The home.jsp page is correct, cuz when I rewrite it to home.html and open it via web browser, it shows correctly. I would really appreciate any help, because I'm a newbie and I'm really confused of all this new stuff![:(]()
Thanks a lot.

Would be anybody please so kind and could show some explicit solution (where to add/modify something and what exactly to add/modify) ?
my app structure in navigator:

Content of web.xml
/************************ web.xml ****************************/ <?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" 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_2_5.xsd"> <!-- The definition of the Root Spring Container shared by all Servlets and Filters --> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/spring/root-context.xml</param-value> </context-param> <!-- Creates the Spring Container shared by all Servlets and Filters --> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <!-- Processes application requests --> <servlet> <servlet-name>appServlet</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>appServlet</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> </web-app>
Content of servlet-context.xml
/************************ servlet-context.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:property name="suffix" value=".jsp" /> </beans:bean> <context:component-scan base-package="com.starter.web" /> </beans:beans>
I'm running my app on tomcat server.
The home.jsp page is correct, cuz when I rewrite it to home.html and open it via web browser, it shows correctly. I would really appreciate any help, because I'm a newbie and I'm really confused of all this new stuff

Thanks a lot.