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.