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<bean id="bestSeller" class="org.springframework.beans.factory.config.PropertyPathFactoryBean"> <property name="targetObject" ref="productRanking" /> <property name="propertyPath" value="bestSeller" /> </bean> </beans> Note that the propertyPath property of PropertyPathFactoryBean can accept not only a single property name, but also a property path with dots as the separators. The preceding bean configuration is equivalent to the following code snippet: Product bestSeller = productRanking.getBestSeller(); In addition to specifying the targetObject and propertyPath properties explicitly, you can combine them as the bean name of PropertyPathFactoryBean. The downside is that your bean name may get rather long and verbose. <bean id="productRanking.bestSeller" class="org.springframework.beans.factory.config.PropertyPathFactoryBean" /> Spring 2.x allows you to declare a bean from an object property or a property path by using the <util:property-path> tag. Compared to using PropertyPathFactoryBean, it is a simpler way of declaring beans from properties. But before this tag can work, you must add the util schema definition to your <beans> root element. <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:util="http://www.springframework.org/schema/util" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-2.5.xsd"> ... <util:property-path id="bestSeller" path="productRanking.bestSeller" /> </beans> You can test this property path by retrieving it from the IoC container and printing it to the console. package com.apress.springrecipes.shop; ... public class Main { public static void main(String[] args) throws Exception { ... Product bestSeller = (Product) context.getBean("bestSeller"); System.out.println(bestSeller); } }

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Applying the Post/Redirect/Get Design Pattern However, when you refresh the web page in the form success view, the form you just submitted will be resubmitted again. This problem is known as duplicate form submission. To avoid this problem, you can apply the post/redirect/get design pattern, which recommends redirecting to another URL after a form submission is handled successfully, instead of returning an HTML page directly. First, you use ParameterizableViewController to define a controller that only renders the reservationSuccess view, which is mapped to reservationSuccess.jsp. <bean id="reservationSuccessController" class="org.springframework.web.servlet.mvc.ParameterizableViewController"> <property name="viewName" value="reservationSuccess" /> </bean> Then you define an explicit mapping for this controller in SimpleUrlHandlerMapping, as ControllerClassNameHandlerMapping won t generate a mapping for a built-in Spring MVC controller: <bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping"> ... <property name="mappings"> <props> ... <prop key="/reservationSuccess.htm">reservationSuccessController</prop> </props> </property> </bean> As you have ResourceBundleViewResolver configured in your web application context, you can define the following redirect view in views.properties in the classpath root: reservationSuccessRedirect.(class)= org.springframework.web.servlet.view.RedirectView reservationSuccessRedirect.url=reservationSuccess.htm Finally, you specify this redirect view for the success view of ReservationFormController. Now when the form submission is handled successfully, the user will be redirected to another URL. Even if the user refreshes this page, it won t cause the duplicate form submission problem. <bean id="reservationFormController" class="com.apress.springrecipes.court.web.ReservationFormController"> <property name="reservationService" ref="reservationService" /> <property name="formView" value="reservationForm" /> <property name="successView" value="reservationSuccessRedirect" /> </bean>

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package com.apress.springenterpriserecipes.post; import javax.ejb.CreateException; import org.springframework.ejb.support.AbstractStatelessSessionBean; public class PostageServiceBean extends AbstractStatelessSessionBean implements PostageService { private PostageService postageService; protected void onEjbCreate() throws CreateException { postageService = (PostageService) getBeanFactory().getBean("postageService"); } public double calculatePostage(String country, double weight) { return postageService.calculatePostage(country, weight); } } When you extend the AbstractStatelessSessionBean class, your EJB class no longer needs to implement any EJB life cycle methods, but you can still override them if necessary. Note that this class has an onEjbCreate() method that you must implement to perform initialization tasks. Here, you just retrieve the postageService bean from the Spring IoC container for this EJB component to use. Of course, you must define it in a bean configuration file. This file can have an arbitrary name, but must be located in the classpath. For example, you can create it as beans-ejb.xml in the root of the classpath.

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Initializing the Command Object You can specify a command class for SimpleFormController, and it will be instantiated for binding form field values. However, in some cases you may have to initialize the command object by yourself. For example, let s consider a situation where you are going to bind a player s name and phone to a Reservation object s player property: <html> <head> <title>Reservation Form</title> </head> <body> <form method="POST"> <table> ... <tr> <td>Player Name</td> <td><form:input path="player.name" /></td> <td><form:errors path="player.name" cssClass="error" /></td> </tr> <tr> <td>Player Phone</td> <td><form:input path="player.phone" /></td> <td><form:errors path="player.phone" cssClass="error" /></td> </tr> <tr> <td colspan="3"><input type="submit" /></td> </tr> </table> </form> </body> </html> However, when the command class Reservation has just been instantiated, the player property is null, so it will cause an exception when rendering this form. To solve this problem, you have to initialize the command object by yourself. You can override the formBackingObject() method of SimpleFormController for this purpose. The default implementation of this method is just to instantiate the command class. When you override this method, the command class is not necessary, as SimpleFormController won t instantiate this class for you any more. package com.apress.springrecipes.court.web; ... public class ReservationFormController extends SimpleFormController { ... public ReservationFormController() { // Don't need to specify the command class // setCommandClass(Reservation.class);

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