Background
Many time we use custom CSS and java script file to add functionality and style to our web pages. We usually refer them in our JSP pages. In this post we will see how we can refer such static resources with Spring 3.0+. We will use Tomcat as our container/server.
Setup
If you are new to Spring then try out a sample Spring program -
Above post is somewhat old but should give you good idea as to get started. Anyhow I will provide the setup xmls here.
Your web.xml file should look as follows -
<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Spring Web MVC Demo Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</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>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/root-context.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
Important file here is
mvc-dispatcher-servlet.xml which forms a part of web application context. Have given a dummy file
root-context.xml for root application context but you may not add it in your web.xml file. If you are not familiar with context loading part I suggest read
As I said create file
mvc-dispatcher-servlet.xml with following content -
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="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 http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/pages/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
<!-- Scan for components under this package -->
<context:component-scan base-package="com.osfg.test" />
<mvc:annotation-driven/>
</beans>
As you can see the web application context will scan package
com.osfg.test to find Spring resources like handler/interceptor. GO ahead create this package and create a controller in it. Lets call it
TestController.java. Add following content to it.
package com.osfg.test;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
/**
* @author athakur
*/
@Controller
public class TestController {
@RequestMapping(value="/test", method=RequestMethod.GET)
public String welcome() {
return "test";
}
}
That's it. We have our controller all setup. This controller returns a logical view name called test. Now if you notice our spring configuration we have already defined
InternalResourceViewResolver which provides a JSTL view. So go ahead and create a JSP file at
/pages/test.jsp under webapp or webcontent.
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>OSFG Test Page</title>
<link href="CSS/test.css" rel="stylesheet">
</head>
<body>
<h1>Hello World from OSFG!</h1>
</body>
</html>
As you can see our JSP references a CSS file called
test.css. Create a folder called CSS and then create a file called test.cs in it. Add following contents to it.
H1 {
color: white; background: teal; FONT-FAMILY: arial; FONT-SIZE: 18pt; FONT-STYLE: normal; FONT-VARIANT: normal
}
Ok we are all set to test now.
Note the Directory structure below -
Testing 1.0
Go ahead start up your tomcat server and load following URL
- http://localhost:8080/TestWebProject/test
Oops! what happened? It is not able to find the CSS file.Well lets come to the point. Spring does not directly allow you to access static resources and the very goal of this post is to show you how. So lets see that now.
Add following to your Spring configuration -
<mvc:default-servlet-handler />
This basically allows you to render static resources from root folder (under webapp or webcontent).
It essentially configures a handler for serving static resources by forwarding to the Servlet container's default Servlet (
DefaultServletHttpRequestHandler is used).
Lets test with our new configuration -
Testing 2.0
After restarting the server hit the URL again -
All good? Yup :) There is also alternate way to do this. You can define resource mapping instead of specifying default-servlet-handler.
<mvc:resources mapping="/resources/**" location="/CSS/" />
Then you also need to make change in JSP as follows
<link href="resources/test.css" rel="stylesheet">
So when you reference resources it will go and pick up from CSS. You can use classpath references in location as well. Here ResourceHttpRequestHandler is used to serve request based on location provided.
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