Spring - handle multiple forms submit with one controller

Like skaffman said you can use @Controller that are very flexible. But if you're writting a Wizard, you should take a look at AbstractWizardFormController. It handles all steps and validation of a wizard. It also keeps the same backing object for each form submission. When your wizard is on the final action, you can take this object and update it at the database.

You can find an exemple at : http://www.cs.bham.ac.uk/~aps/syllabi/2004_2005/issws/h04/spring.html#AbstractWizardFormController

you can use multi action controller .In Spring Single Controller Handel Multipal Method by method name resolver

 <bean id="id" class="controller class"> <property name="name"><ref bean="service Name"/></property > <property
 name="methodNameResolver"><ref bean="name of Resolver"/></property>
 </bean>
 ------------------------------------------------------------------------ <bean id="name of Resolver"
 class="org.springframework.web.servlet.mvc.multiaction.PropertiesMethodNameResolver">
 <property name="mappings"> <props> <prop
 key="/url.com">MethodName</prop> </props> </property> </bean> This can
 be Help you use single controller
     ------------------------------------------------------------------------

https://jira.springsource.org/browse/SPR-5534

Question For example, i have 3 steps registration that map with "/register" url. Is it possible to have only one RegisterController that can handle all registration's steps form submit?

Proposed Solution: Spring 3 - Spring-mvc controller using servlet mapping to your controller. Using the mapping in your url request you will always go to the same controller.

You can create a servlet & servlet mapping that maps to your controller through the spring DispatcherServlet.

Create a servlet & servlet mapping to handle your requests.

Web.XML

    <servlet>
    <description>
    </description>
    <display-name>TestServ</display-name>
    <servlet-name>main</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>test</servlet-name>
    <url-pattern>/test/*</url-pattern>
</servlet-mapping>

Create a spring config file associated to the servlet.

In this case since the name of my servlet is test, it will be test-servlet.xml file that goes into your WEB-INF folder at the root.

<?xml version="1.0" encoding="UTF-8"?>
      <beans xmlns="http://www.springframework.org/schema/beans"
xmlns:p="http://www.springframework.org/schema/p"
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-2.5.xsd">


<bean name="/registration/*" class="register.ResgisterController"/>

Create your controller class to map your requests.

package registration;

import java.util.ArrayList;
import java.util.List;

import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;

import registration.User;

@Controller
public final class RegistrationController {

    public RosterController() {

    }

    //Spring lets you You can access your spring mvc model from the controller automatically 
    @RequestMapping
    public void list(Model model) {
        //do something
    }


    //Extract a registration id from the request and you can use in your model
    @RequestMapping
    public void member(@RequestParam("registrationId") Integer id, Model model) {
        //do something
    }
}

you will have to use 'MultiActionController', but there are some complication in validation porting if you will use this controller.
Solution of similar problem is given @ http://www.scribd.com/doc/20156448/Multi-Action-Controller-With-Validation

you can use Spring webFlow and annotate your controller as @controller