I'm new to jpa and jsf. I'm trying to develop simple web app using jpa and jsf. I have installed mysql and have configured Glassfish to work with it. I have created small project, that consists of one Managed Bean and one JPA Entity. While deployment the table is successfully created in database, so connection to database is ok. But I have one issue, I cant't persist any entity in Managed Bean method:
@ManagedBean(name = "data")
@SessionScoped
public class Data implements Serializable {
@PersistenceUnit(unitName = "JChatPU")
EntityManagerFactory emf;
@Resource
UserTransaction utx;
public Data() {
}
public void add() {
EntityManager em = emf.createEntityManager();
try {
utx.begin();
JChatUser u = new JChatUser();
em.persist(u);
utx.commit();
}
catch(Exception ex) {
ex.printStackTrace();
}
}
}
My persistence unit configuration in persistence.xml:
<persistence-unit name="JChatPU" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>jdbc/jchatdb</jta-data-source>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
<property name="hibernate.hbm2ddl.auto" value="create-drop"/>
</properties>
</persistence-unit>
But I can successfully persist from EJB. So, how to persist from managed bean's method?
Thank's for the future answers.
I have solved the issue!
The problem was in the way I have called bean method:
<h:commandButton id="add" value="add" action="#{data.add()}" />
It doesn't work, because we need to plase command button in the form:
<h:form>
<h:commandButton id="add" action="#{data.meth()}" value="add"/>
</h:form>