How to make a menu in Hibernate + jsf + primefaces?

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How to make a menu in Hibernate + jsf + primefaces?

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#1 by (1 votes)

-1

I just edited the question. Well I already made the menu but without hibernate better said with jdbc. but at the time of deploying the project does not show me the menu with hibernate I will leave the tables.

I'm working with netbeans and when doing with hibernate, they generate the classes in a model package.

now in the dao.

public interface relacionDao { public List listRelation (); }

in the daoImpl

public class relacionDaoImpl implements relacionDao {

@Override
public List<Relacion> listarRelacion() {
   List<Relacion> lista=null;
   Session session=HibernateUtil.getSessionFactory().openSession();
   Transaction t=session.beginTransaction();
    String hql="From Relacion";
    try{
        lista=session.createQuery(hql).list();
        t.commit();
        session.close();
    }catch(Exception e){
        t.rollback();
    }
    return lista;
}

on the bean.

@Named // @ ManagedBean @SessionScoped public class relacionBean implements Serializable {

private List<Relacion> lista;
private Relacion relacion = new Relacion();
private MenuModel model;
private relacionDaoImpl relaciondao = new relacionDaoImpl();

@PostConstruct
public void init() {
    this.listarMenus();
    model = new DefaultMenuModel();
    this.establecerPermisos();
}

public void listarMenus() {

    try {
        lista = relaciondao.listarRelacion();
    } catch (Exception e) {

    }
}

public MenuModel getModel() {
    return model;
}

public void setModel(MenuModel model) {
    this.model = model;
}

public void establecerPermisos() {
    Empleado lg = (Empleado) FacesContext.getCurrentInstance().getExternalContext().getSessionMap().get("empleado");
    for (Relacion m : lista) {
        if (!m.getSubmenu().isProcesado()) {
            int actual = m.getSubmenu().getMenu().getIdMenu();

            if (m.getSubmenu().getMenu().getTipo().equals("S") && m.getEmpleado().getCodEmpleado() == lg.getCodEmpleado()) {
                DefaultSubMenu firstSubMenu = new DefaultSubMenu(m.getSubmenu().getMenu().getNomMenu());

                for (Relacion i : lista) {

                    if (actual == i.getSubmenu().getMenu().getIdMenu() && !i.getSubmenu().isProcesado()) {
                        Submenu submenu = i.getSubmenu();
                        Empleado empleado = i.getEmpleado();
                        if (submenu != null) {
                            if (submenu.getMenu().getIdMenu() == m.getSubmenu().getMenu().getIdMenu() && empleado.getCodEmpleado() == m.getEmpleado().getCodEmpleado()) {
                                DefaultMenuItem item = new DefaultMenuItem(i.getSubmenu().getNombreSubMenu());
                                item.setUrl(i.getSubmenu().getLinkSubMenu());
                                firstSubMenu.addElement(item);
                                i.getSubmenu().setProcesado(true);
                            }

                        }

                    }
                }
                model.addElement(firstSubMenu);
                m.getSubmenu().setProcesado(true);

            } else if (m.getSubmenu().getMenu().getIdMenu() != m.getSubmenu().getIdSubMenu() && m.getEmpleado().getCodEmpleado() == lg.getCodEmpleado()) {
                DefaultMenuItem item = new DefaultMenuItem(m.getSubmenu().getMenu().getNomMenu());
                item.setUrl(m.getSubmenu().getLinkSubMenu());
                model.addElement(item);
            }

        }
    }
}

in the xhtml template.

Facelets Template

<h:body>

    <div id="top" >
        <ui:insert name="top">
            <h:form>
                <p:menubar model="#{relacionBean.model}"/>


            </h:form>

        </ui:insert>
    </div>

    <div id="content" class="center_content">
        <ui:insert name="content">Content</ui:insert>
    </div>

    <div id="bottom">
        <ui:insert name="bottom">Bottom</ui:insert>
    </div>

</h:body>

making the display does not show me the menu. but when I do it with JDBC if it generates the menu I hope you excuse me but a little I do not understand hibernate thank you very much.

java hibernate

asked by Libra2880 24.09.2016 в 03:42

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1 answer

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score 1 · Answer 1

You can not carry the code you have just put because we do not know how your entities are. However, basically Hibernate / JPA works like this.

Each class must be annotated with @Entity or be listed in the persistence unit ( persistence.xml ).
Every entity must have a field annotated with @Id , which represents the primary key.
Relationships between entities are made using @OneToOne , @OneToMany and @ManyToMany .
Entities must implement the Serializable interface.

Example

@Entity
public class Order implements Serializable {

    @Id @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "order_id")
    private Long id;
    @Temporal(TemporalType.TIMESTAMP)
    private Date date;
    private double amount;
    @OneToMany
    @JoinTable(
        name = "order_order_details",
        inverseJoinColumns=@JoinColumn(name = "order_detail_id"),
        joinColumns=@JoinColumn(name = "order_id")
    )
    private List<OrderDetail> details;

    // constructor, getters y setters
}

The entity OrderDetail contains the products and the quantity requested.

@Entity
public class OrderDetail implements Serializable {

    @Id @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "order_detail_id")
    private Long id;
    @OneToOne
    private Product product;
    private short quantity;

    // constructores, getters y setters
}

Entity Producto can be deducted.

Tip: In Hibernate / JPA there is a concept called fetch type or type of procurement , which indicates how we want to obtain relationships strong>. There are 2 types:

Eager (FetchType.EAGER): relations will be obtained together with the main entity. Let's suppose that you get an order, then at that same moment the list of details is brought along with it.
Lazy (FetchType.LAZY): Relationships are brought only when explicitly obtained by means of the getter.

If an order has many details, it is better to make that relationship late.

@OneToMany(fetch = FetchType.LAZY)

In your code I see a considerable amount of Joins making the code slightly legible. With Hibernate / JPA all this is done internally. To get an order with your detail, just call the getter, like this:

Order order = em.find(4359L, Order.class);
List<OrderDetail> details = order.getDetails();

// Imprime el producto y la cantidad de cada detalle 
details.stream().forEach(System.out::println);

Tip: You can see the queries created by Hibernate if you add the following lines to persistence.xml :

<property name="hibernate.show_sql" value="true" />
<property name="hibernate.format_sql" value="true" />

As you can see, JPA / Hibernate abstracts you from many tasks with the database, like any ORM. However, not every project justifies using an ORM; The following must be taken into account:

High performance need
Complexity of the scheme

If we need high performance, it would be best to use a lightweight library such as jOOQ or JDBC flat. If our project is small, then it does not justify using an ORM.