Struts 2 + Spring 2 + JPA + AJAX （三)

JPA configuration

1. Create a folder named "META-INF" under the "src" folder.
2. Create a file named "persistence.xml" under the "META-INF" folder and set its content to:

persistence.xml

<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"
version="1.0">
<persistence-unit name="punit">
</persistence-unit>
</persistence>

JPA configuration can be set on this file. On this example it will be empty because the datasource configuration will be in the Spring configuration file.
Spring

1. Update the content of web.xml under /WebContent/WEB-INF/web.xml to:

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="person" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>person</display-name>

<!-- Include this if you are using Hibernate -->
<filter>
<filter-name>Spring OpenEntityManagerInViewFilter</filter-name>
<filter-class>
org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter
</filter-class>
</filter>

<filter-mapping>
<filter-name>Spring OpenEntityManagerInViewFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>

<filter>
<filter-name>struts2</filter-name>
<filter-class>
org.apache.struts2.dispatcher.FilterDispatcher
</filter-class>
</filter>

<filter-mapping>
<filter-name>struts2</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>


<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>

<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
</web-app>

This will make the container redirect all requests to Struts "FilterDispatcher" class. "index.jsp" is set as the home page, and Spring's "ContextLoaderListener" is configured as a listener.

1. Create a file named "applicationContext.xml" under /WebContent/WEB-INF, and set its content to:

applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-2.0.xsd
http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.0.xsd">

<bean
class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />

<bean id="personService" class="quickstart.service.PersonServiceImpl" />

<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="jpaVendorAdapter">
<bean
class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="database" value="MYSQL" />
<property name="showSql" value="true" />
</bean>
</property>
</bean>

<bean id="dataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://localhost/quickstart" />
<property name="username" value="root" />
<property name="password" value="root" />
</bean>

<bean id="transactionManager"
class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>

<tx:annotation-driven transaction-manager="transactionManager" />

<bean id="personAction" scope="prototype"
class="quickstart.action.PersonAction">
<constructor-arg ref="personService" />
</bean>
</beans>

Note that the "class" attribute of the bean "personAction" is set to the name of the action class, and the "personService" bean will be passed as a parameter to the action constructor. Change the "url", "username" and "password" in the "dataSource" bean to the appropiate values for your database. For more details on the rest of the beans on this file, see Spring's documentation. The "scope" attribute is new in Spring 2, and it means that Spring will create a new PersonAction object every time an object of that type is requested. In Struts 2 a new action object is created to serve each request, that's why we need scope="prototype".
Struts

We will now create a simple Struts action that wraps PersonServices methods, and we will configure Struts to use Spring as the object factory.

1. Open the new class dialog (File -> New -> Class) and enter "PersonAction" for the classname, and "quickstart.action" for the namespace. Set its content to:

PersonAction.java

package quickstart.action;

import java.util.List;

import quickstart.model.Person;
import quickstart.service.PersonService;

import com.opensymphony.xwork2.Action;
import com.opensymphony.xwork2.Preparable;

public class PersonAction implements Preparable {
private PersonService service;
private List<Person> persons;
private Person person;
private Integer id;

public PersonAction(PersonService service) {
this.service = service;
}

public String execute() {
this.persons = service.findAll();
return Action.SUCCESS;
}

public String save() {
this.service.save(person);
this.person = new Person();
return execute();
}

public String remove() {
service.remove(id);
return execute();
}

public List<Person> getPersons() {
return persons;
}

public Integer getId() {
return id;
}

public void setId(Integer id) {
this.id = id;
}

public void prepare() throws Exception {
if (id != null)
person = service.find(id);
}

public Person getPerson() {
return person;
}

public void setPerson(Person person) {
this.person = person;
}
}

Look mom my action is a simple POJO!
The "Preparable" interface instructs Struts to call the "prepare" method if the "PrepareInterceptor" is applied to the action (by default, it is). The constructor of the action takes a "PersonService" as a parameter, which Spring will take care of passing when the action is instatiated.

1. Create a new file named "struts.xml" under the "src" folder. And set its content to:

struts.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<constant name="struts.objectFactory" value="spring" />
<constant name="struts.devMode" value="true" />

<package name="person" extends="struts-default">

<action name="list" method="execute" class="personAction">
<result>pages/list.jsp</result>
<result name="input">pages/list.jsp</result>
</action>

<action name="remove" class="personAction" method="remove">
<result>pages/list.jsp</result>
<result name="input">pages/list.jsp</result>
</action>

<action name="save" class="personAction" method="save">
<result>pages/list.jsp</result>
<result name="input">pages/list.jsp</result>
</action>
</package>

</struts>

Setting "struts.objectFactory" to "spring" will force Struts to instantiate the actions using Spring, injecting all the defined dependencies on applicationContext.xml. The "class" attribute for each action alias is set to "personAction", which is the bean id that we defined on applicationContext.xml for the PersonAction class. This is all that is needed to make Struts work with Spring.