Spring MVC - Parameter Method Name Resolver Example

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The following example shows how to use the Parameter Method Name Resolver of a Multi Action Controller using the Spring Web MVC framework. The MultiActionController class helps to map multiple URLs with their methods in a single controller respectively.

package com.howcodex;

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;
   }

   public ModelAndView add(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Add");
      return model;
   }

   public ModelAndView remove(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Remove");
      return model;
   }
}

<bean class = "com.howcodex.UserController">
   <property name = "methodNameResolver">
      <bean class = "org.springframework.web.servlet.mvc.multiaction.ParameterMethodNameResolver">
         <property name = "paramName" value = "action"/>
      </bean>
   </property>
</bean>

For example, using the above configuration, if URI −

/user/*.htm?action=home is requested, DispatcherServlet will forward the request to the UserController home() method.
/user/*.htm?action=add is requested, DispatcherServlet will forward the request to the UserController add() method.
/user/*.htm?action=remove is requested, DispatcherServlet will forward the request to the UserController remove() method.

To start with, let us have a working Eclipse IDE in place and adhere to the following steps to develop a Dynamic Form based Web Application using the Spring Web Framework.

Step	Description
1	Create a project with a name TestWeb under a package com.howcodex as explained in the Spring MVC - Hello World chapter.
2	Create a Java class UserController under the com.howcodex 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.howcodex;

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;
   }

   public ModelAndView add(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Add");
      return model;
   }

   public ModelAndView remove(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Remove");
      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.mvc.support.ControllerClassNameHandlerMapping"> 
      <property name = "caseSensitive" value = "true" />
   </bean>
   <bean class = "com.howcodex.UserController">
      <property name = "methodNameResolver">
         <bean class = "org.springframework.web.servlet.mvc.multiaction.ParameterMethodNameResolver">
            <property name = "paramName" value = "action"/>
         </bean>
      </property>
   </bean>
</beans>

user.jsp

<%@ page contentType="text/html; charset=UTF-8" %>
<html>
   <head>
      <title>Hello World</title>
   </head>
   <body>
      <h2>${message}</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 the webapps folder using a standard browser. Now, try a URL − http://localhost:8080/TestWeb/user/test.htm?action=home and we will see the following screen, if everything is fine with the Spring Web Application.

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