Marcos Vidolin

Hi guys,

I'm trying to configure Apache Tiles 3 with Spring MVC 3 in my application but the template is not applied.
Does anybody can help me with that configuration?

mvc-dispatcher-servlet.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:context="http://www.springframework.org/schema/context" xmlns:tx="http://www.springframework.org/schema/tx" xmlns:p="http://www.springframework.org/schema/p" xmlns:mvc="http://www.springframework.org/schema/mvc" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.2.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.2.xsd http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.2.xsd"> <mvc:annotation-driven /> <context:annotation-config /> <context:component-scan base-package="br.com.template.controller" /> <bean id="jspViewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver" p:prefix="/WEB-INF/view/" p:suffix=".jsp" /> <mvc:default-servlet-handler /> <mvc:resources mapping="/resources/**" location="/resources/" /> <mvc:view-controller path="/" view-name="index" /> <mvc:view-controller path="/login" view-name="login" /> <mvc:view-controller path="/logout" view-name="logout" /> <mvc:view-controller path="/loginfailed" view-name="login" /> <mvc:view-controller path="/accessdenied" view-name="denied" /> <bean class="org.springframework.web.servlet.view.UrlBasedViewResolver" id="tilesViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.tiles3.TilesView" /> </bean> <bean class="org.springframework.web.servlet.view.tiles3.TilesConfigurer" id="tilesConfigurer"> <property name="definitions"> <list> <value>/WEB-INF/layouts/layouts.xml</value> <value>/WEB-INF/view/**/views.xml</value> </list> </property> </bean> </beans>

layouts.xml
<?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE tiles-definitions PUBLIC "-//Apache Software Foundation//DTD Tiles Configuration 3.0//EN" "http://tiles.apache.org/dtds/tiles-config_3_0.dtd"> <tiles-definitions> <definition name="default" template="/WEB-INF/layouts/default/template.jsp"> <put-attribute name="title" value="Home" /> <put-attribute name="header" value="/WEB-INF/layouts/default/header.jsp" /> <put-attribute name="body" value="" /> <put-attribute name="footer" value="/WEB-INF/layouts/default/footer.jsp" /> </definition> </tiles-definitions>

view.xml
<?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE tiles-definitions PUBLIC "-//Apache Software Foundation//DTD Tiles Configuration 3.0//EN" "http://tiles.apache.org/dtds/tiles-config_3_0.dtd"> <tiles-definitions> <definition name="index" extends="default"> <put-attribute name="body" value="/WEB-INF/view/index.jsp" /> </definition> </tiles-definitions>



Thanks

Hi guys,

I insert the follow line in my entityManagerFactory bean:

<property name="packagesToScan" value="br.com.template.entity" />


Is there any problem doing my web application with entityManagerFactory be a org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean?

[]'s

Hi guys

I'm developing a Spring MVC + Hibernate/JPA project and that is my 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_2_0.xsd" version="2.0"> <persistence-unit name="templatePU"> <provider>org.hibernate.ejb.HibernatePersistence</provider> <class>br.com.template.entity.User</class> <properties> <property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect" /> <property name="hibernate.show_sql" value="true" /> <property name="hibernate.format_sql" value="false" /> </properties> </persistence-unit> </persistence>

Note that I must to declare my entity User (and others ones). I'd like just indicate the entities package like a scan.
How can I do?

Thanks
[]'s

Welcome Adam and enjoy the forum!

[]'s

Hi,

Git is a great idea. All of the most popular open source projects has acceded Git like: jQuery, RichFaces, Rails, SpringFramework... - see https://github.com/popular/forked.
This is the Eclipse plugin that I use: EGit http://www.eclipse.org/egit/

[]'s

These books can help you to get the answers: Effective Java (Joshua Bloch) and Clean Code (Robert C. Martin).
It's a great reading for all Java developers.

[]'s

The approach will depend on what you want. In general I have used JSON because it's smaller and more readable than XML.
Just for note, Twitter and Foursquare left XML and started using exclusively JSON.

The following code snippets represents the same data:


JSON
{ "Student" : [ { "name": "Maria", "grade": [ 8, 9, 7 ] }, { "name": "Peter", "grade": [ 10, 10, 9 ] } ] }

XML
<Students> <Student> <name>Maria</name> <grades> <grade>8</grade> <grade>9</grade> <grade>9</grade> </grades> </Student> <Student> <name>Peter</name> <grades> <grade>10</grade> <grade>10</grade> <grade>9</grade> </grades> </Student> </Students>

Thank's guys

I have a big project with RichFaces 3 (JSF 1) and to migrate RF 3 to 4 its not easy, that's why I need keep project with JSF 1.
But I don't want to get any kind of trouble with that configuration. I'll make some tests to ensure that configuration will work fine.
Do you have some experience with sun-web.xml file?

[]'s

Hi guys,

how can I run JSf 1's applications in Glassfish 3.1 once the application server already implements JSF 2?

tks

Hi Tim Holloway,

I was calling the method using a RichFaces's menu. To resolve that issue I call that same method in a <h:commandLink>, and it's working fine.
* The PDF file it's not showed in the browser but at least I can download it.
I guess something don't work in a ajax request, then if I make a JSP as you said, its will work fine too.

Thank you once again.

Hello,

Guys I need your help once again. =]
I'm trying show a PDF file (using JasperReport) on browser, but it's not working... I have no errors on glassfish's log.
The code was execute completely, but just don't show the report.
That's my code:

@ManagedBean @SessionScoped public class JasperReportsTest { public void printReport() throws JRException, IOException { String path = "/WEB-INF/reports/example.jrxml"; InputStream jasperTemplate = FacesContext.getCurrentInstance() .getExternalContext().getResourceAsStream(path); JasperReport jasperReport = JasperCompileManager .compileReport(jasperTemplate); Map parameters = new HashMap(); JasperPrint jasperPrint = JasperFillManager.fillReport(jasperReport, parameters, new JREmptyDataSource()); HttpServletResponse response = (HttpServletResponse) FacesContext .getCurrentInstance().getExternalContext().getResponse(); response.setContentType("application/pdf"); response.setHeader("Content-Disposition", "attachment; filename=\"report.pdf\""); ServletOutputStream outputStream = response.getOutputStream(); JasperExportManager.exportReportToPdfStream(jasperPrint, outputStream); outputStream.flush(); outputStream.close(); FacesContext.getCurrentInstance().renderResponse(); FacesContext.getCurrentInstance().responseComplete(); } }

Thank you

Hi,

you can do it on Eclipse or NetBeans.
If I understood your question...

[]'s

Hello all,

May I use an @Autowired object from Spring inside my @PostConstruct method?

Tks
[]'s

Hello

Why don't you use the Bean Validation (JSR 303)? Are you using Java EE 6?

[]'s

Thanks Tim,

But I didn't understand how can I use generic DAO like that, injected a EntityManager in all my DAO?

public abstract class AbstractDao<K, E> extends JpaDaoSupport implements IAbstractDao<K, E> { private Class<E> entityClass; public AbstractDao(final EntityManagerFactory factory, final Class<E> entityClass) { setEntityManagerFactory(factory); this.entityClass = entityClass; } public E findById(final K id) { return (E) getJpaTemplate().find(entityClass, id); } public void remove(final E entity) { getJpaTemplate().remove(entity); } public void removeChild(final Object entity) { getJpaTemplate().remove(entity); } ... }