spring mvc hiberate, use CriteriaQuery to run select * from table where - sql

Question

what i need is to run a sql query something like :
select * from table where alpahbetcolumn="A" and numbercolumn="10" and shelfcolumn="upper";
i want to know how to do this query in hibernate using EntityManager
currently this is my own try out, but not working....
#PersistenceContext
EntityManager em;
#Transactional
public List<Item> listItems(String alpahbet, String number, String shelf) {
CriteriaQuery<Item> c = em.getCriteriaBuilder().createQuery(Item.class);
c.from(Item.class);
c..where( em.equal( alpahbet, "alpahbetcolumn" ) && em.equal( number, "numbercolumn" ) && em.equal( shelf, "shelfcolumn" ));
return em.createQuery(c).getResultList();
}
i only have a very vague understanding on spring hibernate topic..still learning...
can someone please point me out how to do this sql query properly, with code example. thanks

Answer 1

Try this
Query q = em.createNativeQuery("select * from table where alpahbetcolumn='A' and numbercolumn= 10 and shelfcolumn='upper'");
q.getResultList();
createNativeQuery() accepts plain SQL as parameter. If you expect Item as result, you can use this
em.createNativeQuery("select * from table where alpahbetcolumn='A' and numbercolumn= 10 and shelfcolumn='upper'", Item.class);
If you want to use JPQL (JPA Query Language), then we need your entity code to be sure, but it would be something like this
em.createQuery("select i from Item i where i.alphabetColumn = 'A' ");
For parameterized queries, use this
Query q = em.createNativeQuery("select * from table where alpahbetcolumn=? and numbercolumn=? and shelfcolumn=?");
q.setParameter(1, "A");
q.setParameter(2, 10);
q.setParameter(3, "upper");

Answer 2

HQL is like this:
StringBuilder queryBuilder = new StringBuilder();
queryBuilder.append(" FROM table");
queryBuilder.append(" WHERE alpahbetcolumn= :codeA");
queryBuilder.append(" AND numbercolumn= :numColumn");
queryBuilder.append(" AND shelfcolumn= :upper");
Query query = getSession().createQuery(queryBuilder.toString());
query = query.setParameter("codeA", "A");
query.setParameter("numColumn", "10");
query.setParameter("upper", "upper");
query.list(); to get your result ;)
and getSession() come from :
private SessionFactory sessionFactory;
#Required
#Autowired
public void setSessionFactory(SessionFactory sessionFactory) {
this.sessionFactory = sessionFactory;
}
protected Session getSession() {
return sessionFactory.getCurrentSession();
}
see spring configuration to configure your sessionfactory ;)
or if you use JPA try this link
get session from entityManager

Answer 3

i have worked out using criteriaquery to fullfill my needs, here is the code. it works
#Transactional
public List<Item> listItems(String alpahbet, String number, String shelf) {
CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
CriteriaQuery<Item> criteriaQuery = criteriaBuilder.createQuery(Item.class);
Root<Item> itemRoot = criteriaQuery.from(Item.class);
criteriaQuery.select(itemRoot).where(criteriaBuilder.equal(itemRoot.get("alpahbetField"), alpahbet), criteriaBuilder.equal(itemRoot.get("numberField"), number), criteriaBuilder.equal(itemRoot.get("shelfField"), shelf));
return em.createQuery(criteriaQuery).getResultList();
}