The following example show how to use Parameterizable View Controller
method of a Multi Action Controller using Spring Web MVC framework.
Parameterizable View allows to map a web-page with a request.
UserController.java
Now start your Tomcat server and make sure you are able to access other web pages from webapps folder using a standard browser. Now try a URL http://localhost:8080/TestWeb/index.htm and you should see the following result if everything is fine with your Spring Web Application:
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 above configuration, if URI
- /index.htm is requested, DispatcherServlet will forward the request to the UserController controller with viewName set as user.jsp.
| Step | Description |
|---|---|
| 1 | Create a project with a name TestWeb under a package com.tutorialspoint as explained in the Spring MVC - Hello World Example chapter. |
| 2 | Create a Java classes UserController under the com.tutorialspoint package. |
| 3 | Create a view file user.jsp under jsp sub-folder. |
| 4 | The final step is to create the content of all the source and configuration files and export the application as explained below. |
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 and use Export > WAR File option and save your TestWeb.war file in Tomcat's webapps folder.
Now start your Tomcat server and make sure you are able to access other web pages from webapps folder using a standard browser. Now try a URL http://localhost:8080/TestWeb/index.htm and you should see the following result if everything is fine with your Spring Web Application:
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