How to retrieve only certain fields of an entity in JPQL or HQL? What is the equivalent of ResultSet in JPQL or HQL?

Question

In JPQL, I can retrieve entities by :

query = entityManager.createQuery(\"select c from Category c\");
List categories = query.getResultList(         


        

Answer 1

It is not the use of the .list() function itself which makes the result a List<Object[]>. It is the specification of fields (c.id, c.name) in the HQL query. If your query is

    "select c from Category c"

Then query.list() will return a List<Category> object.

Answer 2

You can also directly map to the class

public class UserData {

    private String name;
    private Date dob;
    private String password;
//setter
}



  public UserData getUserData() {
        String queryString = "select user.name as name, user.dob as dob, user.userses.password as password from UserProfile user where user.userEmailId='faiz.krm@gmail.com'";
        Query query = sessionFactory.getCurrentSession().createQuery(queryString);
        query.setResultTransformer(Transformers.aliasToBean(UserData.class));
        return query.uniqueResult();
    }

Answer 3

In HQL you can use list() function to get a list of Object[] array that contains result rows:

Query query = session.createQuery("select c.id,c.name from Category c");
List<Object[]> rows = query.list();

in returned array 1-st element will be id, second - name.

for (Object[] row: rows) {
    System.out.println(" ------------------- ");
    System.out.println("id: " + row[0]);
    System.out.println("name: " + row[1]);
}

If you want to use hibernate's Criteria API, you should use Projections.

With JPA it will work the same way:

List<Object[]> rows = entityManager.createQuery(queryString).getResultList();

Answer 4

The JPA specification allows us to customize results in an object-oriented fashion.

More information you can find at this link

There is an even more elegant way