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The following example shows how to use the Parameterizable View Controller method of a Multi Action Controller using the Spring Web MVC framework. The Parameterizable View allows mapping a webpage with a request.
package com.tutorialspoint;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;
public class UserController extends MultiActionController{
public ModelAndView home(HttpServletRequest request,
HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("user");
model.addObject("message", "Home");
return model;
}
}
<bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<value>
index.htm=userController
</value>
</property>
</bean>
<bean id="userController" class="org.springframework.web.servlet.mvc.ParameterizableViewController">
<property name="viewName" value="user"/>
</bean>
For example, using the above configuration, if URI.
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/index.htm is requested, DispatcherServlet will forward the request to the UserController controller with viewName set as user.jsp.
To start with it, let us have a working Eclipse IDE in place and stick to the following steps to develop a Dynamic Form based Web Application using Spring Web Framework.
| Step | Description |
|---|---|
| 1 | Create a project with a name TestWeb under a package com.tutorialspoint as explained in the Spring MVC – Hello World chapter. |
| 2 | Create a Java class UserController under the com.tutorialspoint package. |
| 3 | Create a view file user.jsp under the jsp sub-folder. |
| 4 | The final step is to create the content of the source and configuration files and export the application as explained below. |
UserController.java
package com.tutorialspoint;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;
public class UserController extends MultiActionController{
public ModelAndView home(HttpServletRequest request,
HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("user");
model.addObject("message", "Home");
return model;
}
}
TestWeb-servlet.xml
<beans xmlns = "http://www.springframework.org/schema/beans"
xmlns:context = "http://www.springframework.org/schema/context"
xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation = "
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<bean class = "org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name = "prefix" value = "/WEB-INF/jsp/"/>
<property name = "suffix" value = ".jsp"/>
</bean>
<bean class = "org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name = "mappings">
<value>
index.htm = userController
</value>
</property>
</bean>
<bean id = "userController" class = "org.springframework.web.servlet.mvc.ParameterizableViewController">
<property name = "viewName" value="user"/>
</bean>
</beans>
user.jsp
<%@ page contentType="text/html; charset=UTF-8" %>
<html>
<head>
<title>Hello World</title>
</head>
<body>
<h2>Hello World</h2>
</body>
</html>
Once you are done with creating source and configuration files, export your application. Right click on your application, use Export → WAR File option and save the TestWeb.war file in Tomcat”s webapps folder.
Now, start your Tomcat server and make sure you are able to access other webpages from webapps folder using a standard browser. Now, try a URL – http://localhost:8080/TestWeb/index.htm and you will see the following screen, if everything is fine with the Spring Web Application.
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