在Facade模式中放置用于创建命名查询的代码的位置?

时间:2014-06-12 13:01:05

标签: java gwt jpa ejb facade

我正在为我的应用程序使用GWT,Oracle数据库,Glassfish服务器,JPA。在客户端没有问题。还在服务器端使用Facade模式。

AbstractFacade.java

public abstract class AbstractFacade<T> {
private Class<T> entityClass;

public AbstractFacade(Class<T> entityClass) {
    this.entityClass = entityClass;
}

protected abstract EntityManager getEntityManager();

public void create(T entity) {
    getEntityManager().persist(entity);
}

public void edit(T entity) {
    getEntityManager().merge(entity);
}

public void remove(T entity) {
    getEntityManager().remove(getEntityManager().merge(entity));
}

public T find(Object id) {
    return getEntityManager().find(entityClass, id);
}

public List<T> findAll() {
    javax.persistence.criteria.CriteriaQuery cq = getEntityManager().getCriteriaBuilder().createQuery();
    cq.select(cq.from(entityClass));
    return getEntityManager().createQuery(cq).getResultList();
}

public List<T> findRange(int[] range) {
    javax.persistence.criteria.CriteriaQuery cq = getEntityManager().getCriteriaBuilder().createQuery();
    cq.select(cq.from(entityClass));
    javax.persistence.Query q = getEntityManager().createQuery(cq);
    q.setMaxResults(range[1] - range[0] + 1);
    q.setFirstResult(range[0]);
    return q.getResultList();
}

public int count() {
    javax.persistence.criteria.CriteriaQuery cq = getEntityManager().getCriteriaBuilder().createQuery();
    javax.persistence.criteria.Root<T> rt = cq.from(entityClass);
    cq.select(getEntityManager().getCriteriaBuilder().count(rt));
    javax.persistence.Query q = getEntityManager().createQuery(cq);
    return ((Long) q.getSingleResult()).intValue();
}

GroupFacade.java

@Stateless
public class GroupsFacade extends AbstractFacade<Groups> implements GroupsFacadeLocal {
    @PersistenceContext(unitName = "exam_ejb_1.0PU")
    private EntityManager em;

    @Override
    protected EntityManager getEntityManager() {
        return em;
    }

    public GroupsFacade() {
        super(Groups.class);
    }
}

GroupsFacadeLocal.java

@Local
public interface GroupsFacadeLocal {

    void create(Groups groups);

    void edit(Groups groups);

    void remove(Groups groups);

    Groups find(Object id);

    List<Groups> findAll();

    List<Groups> findRange(int[] range);

    int count();

}

Groups.java

@Entity
@Table(name = "GROUPS")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "Groups.findAll", query = "SELECT a FROM AtsTestJautajumsGrupas a"),
    @NamedQuery(name = "Grous.findByTestId", query = "SELECT a FROM Groups a WHERE a.TestId = :TestId")})
public class Groups implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @NotNull
    @Column(name = "GROUP_ID")
    private BigDecimal groupId;
    @Size(max = 200)
    @Column(name = "name")
    private String name;
    @Column(name = "ARCHIVE_STAT")
    private String archiveStat;
    @OneToMany(mappedBy = "GroupId")
    private Collection<Question> questionCollection;
    @JoinColumn(name = "TEST_ID", referencedColumnName = "TEST_ID")
    @ManyToOne
    private Test atsTestId;

    public Groups() {
    }

    //Getters and Setter here...
}

所以我在创建或查找db中的所有记录时没有任何问题。

 public class ServiceImpl extends RemoteServiceServlet implements Service {
        @EJB
            AtsTestJautajumsGrupasFacadeLocal grupasEJB;
        @Override
            public List<Groups> getCategories() {

                List<Groups> grupas = new ArrayList<Groups>();
                grupas = grupasEJB.findAll();

                return grupas;
            }

但现在我需要选择具有相同Test_id的组。我创建了@NamedQuery(name = "Grous.findByTestId"...,现在我有点迷失了。

所以我的问题是:我如何使用NamedQuery?我在哪里创建一个访问db的方法。我需要从ServiceImpl访问该方法。什么是正确的方法呢?

编辑: 还有一个问题。何时以及为什么我应该使用CriteriaBuilder而不是你建议的?示例:

public AtsTestJautajums getJautajums(Integer atsTestJautajumsId) {
        CriteriaBuilder cb = em.getCriteriaBuilder();
        CriteriaQuery<AtsTestJautajums> q = cb.createQuery(AtsTestJautajums.class);
        Root<AtsTestJautajums> atsTestJautajums = q.from(AtsTestJautajums.class);
        List<Predicate> predicateList = new ArrayList<Predicate>();
        Predicate atsTestJautajumsIdPredicate;
        if (atsTestJautajumsId != null) {
            atsTestJautajumsIdPredicate = cb.equal(atsTestJautajums.<String>get("atsTestJautajumsId"), atsTestJautajumsId);
            predicateList.add(atsTestJautajumsIdPredicate);
        }
        Predicate[] predicates = new Predicate[predicateList.size()];
        predicateList.toArray(predicates);
        q.where(predicates);
        TypedQuery<AtsTestJautajums> query = em.createQuery(q);
        return query.getSingleResult();
    }

1 个答案:

答案 0 :(得分:1)

我不确切知道您的项目如何工作以及您如何编写代码,但基本上我认为以下示例包含您需要执行的所有操作。

public List<Groups> getCategoriesByTestId(Test test){
   return em().createNamedQuery("Grous.findByTestId", Groups .class).setParameter("TestId", test.getId()).getResultList();
}

您的方法应该在您使用的任何服务中调用(在您的代码中似乎是ServiceImpl),并在DAO中实现您的方法,该方法使用jpa处理所有数据库访问(在您的代码中似乎是GroupsFacade)。别忘了在GroupsFacadeLocal中声明你的方法原型。如果我理解你的代码,我认为这就是它。